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Chemical 
Calculations 



BY 



E. HARMAN ASHLEY, Ph.D. 

ASSISTANT PROFESSOR OF CHEMISTRY IN THE UNIVERSITY 
OF MAINE 



ILL USTBA TED 




NEW YORK 

D. VAN NOSTRAND COMPANY 

25 Park Place 

1915 



oX" 



Copyright, 1915> 

BY 

D. VAN NOSTRAND COMPANY 



V 

Stanbope iprcss 

F. H.GILSON COMPANY 
BOSTON, U.S.A. 

©CI,A406817 
JUL 23 1915 



PREFACE 

In putting forth this small work, the author has in- 
tended to write a textbook on Chemical Calculations fol- 
lowing more closely than other books at present on the 
market, the needs of the student who will later find 
occupation in chemical laboratory work. For this reason, 
constant reference will be made to Chemists' Handbooks, 
the best of which, in the author's opinion, is "Van Nos- 
trand's Chemical Annual for 1913," edited by Prof. 
J. C. Olsen. Constant reference will be made to this 
work by means of footnotes, with the intention of lead- 
ing the student to the most convenient source of tabu- 
lated data, and to explain the method of utilizing the 
same= The purchase of the ''Annual" will not be nec- 
essary, much as it may be desirable, in following the ex- 
planations or solving the problems. 

The problems are numerous and of varying degrees of 
difficulty; the instructor is the best judge of the type 
of problem to assign a class or individual. Those of the 
more difficult sort may be given the more advanced 
student, thus holding the class together. Many of the 
problems conform to the type usually found in works of 
this nature and in the selection of these, most of the 
books on this subject have been consulted. Some of the 
types are new to works of this kind; their inclusion is 
made in the hope of making the book more practical. 

The method of solving the problems is treated in the 
text, and in the earlier chapters typical problems are 
solved at the end of the reading matter just before the 
problems are stated. The answer or answers have been 

iii 



IV PREFACE 

attached to each problem rather than given in a list in 
the appendix, with the belief that it is more convenient 
for student and instructor. It may be urged that the 
inclusion of the answer with the problem may be a temp- 
tation offered the student to work for this answer. How- 
ever, a student who will do this, will always take the 
trouble of looking up the answer, even if it is at the end 
of the book. 

Some of the topics are discussed from two or more 
methods of approach in the belief that one of the methods 
may appeal to a student and be more easily comprehended 
by him than the others. If some of the topics are treated 
in too elementary or extended a fashion to suit some, it 
must be remembered that this subject is not simple to all 
students. The author does not anticipate adverse criti- 
cism on the score of the work being too elementary when 
taken as a whole. The treatment of ratios may be so 
full as to be tedious. This is for the reason that it is 
wished to lead the student away from proportions as 
generally expressed, i.e., x : y = a :b, but to have him 
at once set down his mathematical expression in a form 
for immediate solution. The algebraic method has been 
extensively employed and the treatment of indirect 
analysis and the calculation of indirect factors have been 
given considerable space, an amount of space not meas- 
ured by the importance of indirect analysis, but rather 
by the importance of the methods of chemical and alge- 
braic reasoning. Many problems have been solved in 
steps, after which it is shown how the same problem 
may be solved by one expression. The problems are 
more numerous than will be solved by one class, if small; 
two or more problems of the same nature are included 
so that in a large class there will be enough to go around. 
Furthermore, a large number of problems will give the 



PREFACE V 

instructor more choice, and there is less likeUhood of 
their being solved by one class and handed on to the 
next in the nature of heirlooms. 

In the calculation of the problems, the use of loga- 
rithms is strongly recommended; as also is the slide rule. 
The latter is to be especially recommended, as the time 
allotted the subject may be devoted to theory, relegating 
the mechanical solution of the problems to a subordinate 
place. With the use of this device the subject may be 
thoroughly treated in half the usual time. 

The problems have been checked and rechecked, but 
notwithstanding this, errors will undoubtedly creep in. 
The author will appreciate having his attention called to 
such. As five- and, in a few instances, six-place logarithms 
have been used, answers are often stated to a number of 
significant figures not justified by the number of signifi- 
cant figures in the data given. It was thought best to 
carry out the answer, as this involved no extra work and 
is valuable in inculcating accuracy. 

The author wishes to express his appreciation to Mr. 
Harry R. Lee, Superintendent, of the Virginia Electro- 
lytic Company for checking problems, reading the 
manuscript and giving much criticism and advice; to 
Mr. G. C. Merrill, formerly of the College of Montana, 
for the checking of a large number of problems. Profes- 
sors Ralph H. McKee and Charles W. Easley of the 
University of Maine have been generous with help and 
advice, as have many others of the author's colleagues. 
An expression of thanks is due Mr. Charles F. Guhlmann 
of the General Chemical Company for reading the chap- 
ter on mixed acids and offering very helpful advice for 

this section. ^ ^^ . 

R. H. A. 

University of Maine, 

Orono, Maine, 

Jan. 30, 1915. 



TABLE OF CONTENTS 

Page 
Preface '. iii-v 

CHAPTER I: Ratios 1-25 

Decomposition of Mercuric Oxide — Ratios Involved 
in the Decomposition of Mercuric Oxide — Metathesis 
between Barium Chloride and Sodium Sulphate — Fac- 
tors — Percentage Composition of Chemical Com- 
pounds — Equivalents — Successive Reactions — Fac- 
tors Other than Chemical — Problems. 

CHAPTER II: Approximate Numbers 2&-31 

Addition and Subtraction of Approximate Numbers — 
Multiplication of Approximate Numbers — Division 
of Approximate Numbers — Recapitulation. 

CHAPTER III: Interpolation 32-47 

. Functions — Graphic Representation of Functions — 
Change of Percentage Composition with Baume for 
Sulphuric Acid — Interpolation — Extrapolation — 
Problems. 

CHAPTER IV: Heat 48-58 

Fahrenheit Scale — Centigrade Scale — Reaumur 
Scale — Absolute Scale — Calorie — British Thermal 
Unit — Specific Heat — Heat of Fusion — Heat of 
Vaporization — Heat of Combustion — Conversion of 
the Centigrade into British System and Vice Versa — 
Problems. 

CHAPTER V: Specific Gravity 59-79 

Definitions — Specific Gravity of Body Heavier than 
and Insoluble in Water — Specific Gravity of Solid 
Substance Lighter than Water — Specific Gravity of 

vii 



VUl CONTENTS 

Page 

Powders — Pyknometer Method for Liquids — Sinker 
Method for Liquids — Constant Weight Hydrometers 
— Constant Volume Hydrometers — Hydrometers — ■ 
Baume Hydrometer — Conversion of Baume Readings 
into Specific Gravity and Vice Versa — Calculation of 
Density Determinations to Different Conditions of 
Temperature — Problems. 



CHAPTER VI: Gas Calculations 80-140 

Boyle's Law — Charles' Law — Gas Thermometer — ■ 
Law of Boyle and Charles Combined — Standard Con- 
ditions of Temperature and Pressure — Correction of 
Barometer for Temperature — Moist Gases — Gay- 
Lussac's Law — Gas Analysis — Avogadro's Law — 
Atomic Weights and Molecular Weights — Vapor 
Density — Graham's Law — Volume Occupied by a 
Gram Molecule of a Gas — Measurement of Vapor 
Density — Deviations from Gas Laws — Volume of 
Gas from Weight of Substance (Metric System) — 
Volume of Gas from Weight of Substance (English 
System) — Calculations of Weighings in Air to Value 
in Vacuo and Vice Versa — Standard Unit of Volu- 
metric Apparatus — Mohr's Liter — Dalton's Law of 
Partial Pressures — Problems. 



CHAPTER VII: Calculation op Atomic Weights 

AND Formulas 141-171 

Atomic Weights — Valence — Combining Weights — 
Relationship between Atomic Weight, Valence and 
Combining Weight — Calculation of Atomic Weight 
from an Analysis — Law of Dulong and Petit — Molec- 
ular Weight by Elevation of Boiling Point and De- \ 
pression of Freezing Point — Molecular Weight of an 
Organic Acid — Molecular Weight of an Organic Base 
— Formula of a Compound Given Molecular Weight and 
Percentage Composition — Formula from Percentage 
Composition — Formulas of Minerals, Isomorphic 
Replacement — Problems. 



CONTENTS IX 

Page 

CHAPTER ,VIII: Gravimetric Analysis 173-201 

Direct Gravimetric Analysis — Elimination of a Con- 
stituent from an Analysis — Factor Weights — In- 
direct Analysis — Analysis of Oleum — Problems. 

CHAPTER IX: Volumetric Analysis 202-251 

Single Factor Solutions — Methods of Standardization 

— Equivalent Values of Single Factor Solutions — 
Factor Weights for Volumetric Solutions — Normal 
Solutions — Simplification of Calculations by the Use 
of Normal Solutions — Calculation of Normality — • 
Factor to a Given Normality — Volumetric Determi- 
nations using two Solutions — Back Titrations — Vol- 
umetric Analysis using Two Indicators — Volumetric 
Analysis of Oleum — Adjustment of Strength of Solu- 
tions — Problems. 

CHAPTER X: Use of Specific Gravity Tables and 

Acid Calculations 252-267 

Calculation of Data in Specific Gravity Tables — Use 
of Specific Gravity Tables — Calculation of Mixed 
Acid — Strengthening of Mixed Acid by Use of Oleum 

— Problems. 



CHEMICAL CALCULATIONS 



CHAPTER I 
RATIOS 

Chemical calculations are based on the two following 
laws: 

Law of the Conservation of Mass. — In any system un- 
dergoing change, chemical dr otherwise, no change in 
the mass of the system has ever been observed. 

Law of Constancy of Composition. — The masses of the 
elements taking part in a given chemical change always 
exhibit a definite and invariable ratio to each other. 

The Decomposition of Mercuric Oxide. — Let it be 
required to find the number of grams of oxygen which 
will be liberated by the decomposition of 1.7000 g. of 
mercuric oxide. The decomposition is represented by the 
equation: 

2HgO = 2Hg + O2 
parts by weight, 2 (216.6) 2 (200.6) 2 (16) 

In the atomic weight tables,^ 200.6 is given as the atomic 
weight of mercury and 16.00 as that of oxygen. The sym- 
bol- of mercuric oxide, HgO, shows that one molecule of 
mercuric oxide contains one atomic weight of mercury 

1 The atomic weights are revised annually by an international 
committee. See "Van Nostrand's Chemical Annual," p. 1, third 
issue, 1913. Frequent reference will be made to this book, where 
the title will be abbreviated to Chem. Ann., always referring to the 
edition of 1913. 

1 



2 CHEMICAL CALCULATIONS 

and one atomic weight of oxygen, consequently the sum 
of these atomic weights (200.6 + 16.00), 216.60, represents 
the molecular weight of mercuric oxide. ^ Single atoms 
and molecules are so small that they cannot be dealt 
with as such, but aggregations of atoms or molecules may 
be taken in the ratio of their atomic or molecular weights, 
in which case an equal number of atoms are dealt with. 
Numbers representing molecular or atomic weights are 
relative only, being based upon a convenient standard, 
for example, the weight of an atom of oxygen, the atomic 
weight of which is arbitrarily fixed at 16.00. In a complete 
reaction, when all the initial substances are adjusted in 
the ratio of the molecular or atomic weights, all the initial 
substances disappear and the products stand in relation 
to each other in the ratio of their molecular or atomic 
weights and nothing remains over. 

Ratios Involved in the Decomposition of Mercuric 
Oxide. — Returning to the problem, the ratio of oxygen 
to mercuric oxide is 2 (16) parts by weight of oxygen to 

2 (216.6) parts by weight of mercuric oxide, or . 

Zi yZiiXiXi) 

This is the fixed and invariable ratio which always holds 
between these two substances in this particular reaction. 
In other words, 2 (16) parts by weight of oxygen are evolved 
by the decomposition of 2 (216.6) parts by weight of mer- 
curic oxide. Then, if 2 (216.6) parts by weight of mercuric 
oxide yield 2 (16) parts by weight of oxygen, one part by 

weight of oxygen is given by . . parts by weight of 

1 Tabulations of molecular weights are to be found in chemist's 
handbooks. In such a simple case as this, no time would be 
saved by so doing, but in many cases such tabulations are useful 
and much time saved and liability to error reduced. See Chem. 
Ann., pp. 39-47; 100-336. 



RATIOS 3 

mercuric oxide, and as the unit of weight under considera- 
tion is the gram, the problem is solved by the expression: 

2^^^ X 1.7000 = 0.1256 g. of oxygen, yielded by 
1.7000 g. mercuric oxide. 

Again, employing the same equation, how much mer- 
cury will be obtained by the decomposition of 1.7000 g. of 
mercuric oxide? The quantity of mercury is now sought; 
consequently the ratio of mercury to mercuric oxide being 

2(200.6) , . • 1 • +u 

. ' , and reasonmg precisely as m the previous 

problem, the expression becomes: 

I ^l^f^. X 1.7000 = 1.5744 g. of mercury, yield from 
1.7000 g. mercuric oxide. 

In all, this reaction yields the following ratios: 

(ay Oxygen to mercuric oxide, . • 

.7,^ f^ , 2(16) 

(b) Oxygen to mercury, ^ - 



(c) Mercury to mercuric oxide. 



2 (200.6) 
2(216.6)' 



1 It is evident that , . , etc., is equivalent to ; but 

the equation as written is more accurate, chemically speaking, than 
the simpler equation, HgO = Hg + O, in which the oxygen is rep- 
resented as in the atomic state, a condition fulfilled only at the 
instant of decomposition. It is necessary for the student to bear 
this in mind, as in gas equations involving volumes it is a matter 
of importance. For calculations involving weights only, such an 
equation serves the purpose, but in the calculation it is a simple 
matter to write the correct equation, and in solving to cancel; i.e., 

:2f(lQ) ^ 16 
7(216.6) 216.6' 



4 CHEMICAL CALCULATIONS 

2 (200.6) 



(d) Mercury to oxygen, 

(e) Mercuric oxide to oxygen, 
(/) Mercuric oxide to mercury, 



2(16) 
2 (216.6) 

2(16) 
2 (216.6) 



2 (200.6) 
If X represents the units of weight in any system: 

(a)' . . X = weight of O2 produced by x units of HgO. 

^^^^ 9 f9C\c r^ ^ "^ weight of O2 equivalent to x units of Hg. 
(c)' ^':^ x = weight of Hg produced by x units HgO. 
(d)' ^ , ' X = weight of Hg equivalent to x units of O2. 
(e)' ^ \ X = weight of HgO produced hjx units of O2. 
(f)'7^i^X)cr^ X = weight of HgO produced by x units of Hg. 

These six ratios are applied in problems typified by 
the folio wing: 

(a)'' How many grams of oxygen will be liberated by 
the decomposition of 3.000 g. of mercuric oxide? 

Ans. 2^|J|^^ X 3.000 = 0.2216 g. 

{h)" How many grams of oxygen are associated with 
3.000 g. of mercury in mercuric oxide? 

Ans. 2^^^ X 3.000 = 0.2393 g. 



KATIOS 5 

(c)" How many grams of mercury are produced by the 
decomposition of 3.000 g. of mercuric oxide? 

^-- HSfx 3-000 = 2.7784 g. 

{dy How many grams of mercury are associated with 
3.000 g. of oxygen in mercuric oxide? 

Arts. ^^^^ X 3.000 = 37.613 g. 

{e)" How many grams of mercuric oxide must be de- 
composed to obtain 3.000 g. of oxygen? 

Ans. ^^ip X 3.000 = 40.613 g. 

{jY' How many grams of mercuric oxide must be de- 
composed to obtain 3.000 g. of mercury? 

Metathesis between Barium Chloride and Sodium 
Sulphate. — Again take the equation: 

BaCl2 + Na2S04 = BaS04 + 2 NaCl. 

The molecular weights being the sum of the atomic 
weights, are: 

BaCl2 Na2S04 

Ba = 137.37 Nag = 2 (23.00) = 46.00 

2 CI = 2 (35.46) = 70.92 S = 32.07 

208.29 4 0=4 (16.00) = 64.00 

142.07 

BaS04 2 NaCl 

Ba = 137.37 Na = 23.00 

S = 32.07 CI = 35.46 

4 0=4 (16.00) = 64.00 2 (58.46) 

233.44 



b CHEMICAL CALCULATIONS 

As before the ratios are: 

208 29 
a) Barium chloride to sodium sulphate : 



h) Barium chloride to barium sulphate: 

c) Barium chloride to sodium chloride: 

d) Sodium sulphate to barium chloride: 

e) Sodium sulphate to barium sulphate : 
/) Sodium sulphate to sodium chloride: 
g) Barium sulphate to barium chloride: 
h) Barium sulphate to sodium sulphate : 
i) Barium sulphate to sodium chloride: 
j) Sodium chloride to barium chloride: 
k) Sodium chloride to sodium sulphate: 
I) Sodium chloride to barium sulphate: 



142.07 
208.29 
233.44* 
208.29 

2(58.46) 

142.07 

208.29* 

142.07 

233.44* 

142.07 
2 (58.46) 
233.44 
208.29* 
233.44 
142.07* 

233.44 
2 (58.46) 
2 (58.46) 

208.29 
2 (58.46) 

142.07 
2 (58.46) 



233.44 

These twelve ratios cover all cases which will arise 
from this reaction. In the typical examples given below, 
these ratios are employed in solving problems. 

{ay How many grams of barium chloride will be re- 
quired to react with (are equivalent to) 5.0000 g. of sodium 
sulphate? 

Ans. ^^ X 5.0000 = 7.3305 g. 



RATIOS 7 

(by How many grams of barium chloride are required 

for the precipitation of 5.0000 g. of barium sulphate? 

one OQ 
Ans. 2^^ X 5.0000 = 4.4613 g. 

(c)' Required the number of grams of barium chloride 
equivalent to 5.0000 g. of sodium chloride. 

Ans. 2^1^ X 5.0000 = 8.9074 g. 

(d)' How many grams of sodium sulphate are necessary 

for the precipitation of the barium of 5.0000 g. of barium 

chloride? 

142 07 
Ans, 208!^ X 5.0000 = 3.4104 g. 

(e)' How many grams of sodium sulphate have been 

added to barium chloride if 5.0000 g. of barium sulphate 

are precipitated? 

142 07 
^^^- ^^^^ X ^-0000 = 3.0430 g. 

(/)' How many grams of sodium sulphate are equiva- 
lent to 5.0000 g. of sodium chloride? 

142 07 
Ans. 2(5sm ^ ^'^^^^ ^ ^'^^^^ ^' 

((/)' How many grams of barium sulphate are precipi- 
tated by 5.0000 g. of barium chloride? 

Ans. 1^1^ X 5.0000 - 5.6037 g. 

(h)' How many grams of barium sulphate are precipi- 
tated by 5.0000 g. of sodium sulphate? 

OQQ 44 

Ans. ^11^ X 5.0000 = 8.2157 g. 



8 CHEMICAL CALCULATIONS 

(^y How many grams of barium sulphate are equiva- 
lent to 5.0000 g. of sodium chloride? 

Ans. 2^11^ X 5.0000 = 9.9829 g. 

(j)' How many grams of sodium chloride are formed 
when 5.0000 g. of barium chloride are added to a solution 
containing sodium sulphate? 

Ans. ^Qs'y X 5.0000 - 2.8067 g. 

(/c)' How many grams. of sodium chloride are equiva- 
lent to 5.0000 g. of sodium sulphate? 

Ans. ^^^^ X 5.0000 - 4.1149 g. 

(l)' If 5.0000 g. of barium sulphate are precipitated, 
how many grams of sodium chloride are formed? 

Ans. ^^^I'lf X 5.0000 = 2.5043 g.^ 

* These examples are typical of the commonest and simplest 
chemical calculations. It is to be noted that there are three terms, 
of which two are expressed in similar units, the third being dissimi- 
lar. In these examples there are two terms expressing atomic 
weights and one term in grams. To solve a problem of this type: 
Make a fraction of the two terms in similar units, placing 

THAT term representing THE MAGNITUDE SOUGHT OVER THE LINE 
(making it the numerator), the OTHER TERM OF THE SAME KIND 
OF UNIT UNDER THE LINE (MAKING IT THE DENOMINATOR), MULTIPLY 
BY THE TERM OF THE DISSIMILAR UNIT. 

This rule of constructing ratios is, as stated, empirical, but a 
little reflection will show its validity. It is given in this form in 
the belief that it is more easily grasped and that a little practice 
will result in making application of it almost mechanical. It must 
be borne in mind that it is applicable only when an increase in one 
term produces a corresponding increase in the other; in other words, 
in those cases only in which the two units going to make up the 
fraction vary directly. The great majority of chemical problem 



RATIOS 9 

Factors. — The fractions composed of the terms of 
similar units are constant for the same equation and when 
frequently employed it is advantageous to solve the ratio 
once for all. The values thus obtained are called factors.^ 
In the reaction between barium chloride and sodium sul- 
phate, the fractions formed by the molecular weights 
yield the following constants: 

W ^r^ = ttI^ =1.4661: factor of BaCl2 to Na2S04. 

^ ^ Na2S04 142.07 

(by ^^ = I?!?? =0.8923: factor of BaCla to BaS04. 

W ^^^1 = o^/^o ^1 =1.7815: factor of BaCl2 to NaCl. 
2 NaCl 2 (58.46) 

{dy ^J^f?' = l^^= 0.6821 : factor of Na2S04 to BaCla. 
±5aUi2 Zkjo.Zv 

(eV ^^f?' = ill^ =0.6086: factor of Na2S04 to BaS04. 

lDabU4 iiOO.44: 

(hy ^^^ = ??I4I =1.6431 : factor of BaS04 to Na2S04. 
^ ^ Na2S04 142.07 

come under this head, the chief exceptions being in gas calculations, 
as for instance Boyle's law, which states that the volume of a gas 
varies inversely with the pressure. In such cases the rule is re- 
versed so far as the composition of the fraction is concerned. This 
is treated in another place. 

^ See Chem. Ann., pp. 10-36. 

2 When chemical symbols are written in this manner, the frac- 
tion is considered as made up of the terms corresponding to the 

molecular weight of the symbol, ^ ^^ representing the fraction 

208.29 
142.07 • 



10 CHEMICAL CALCULATIONS 

®" m = 2lSS) =l-'^966:factorofBaSO,toNaCl. 
^y ro = M^ ='■''''■■ factorofNaCltoBaa. 
W'lr^ = ?r||47^ = 0.8230: factor to NaCl to NaaSOi. 

W ^l^^i = ^^^ = 0.5009 :factorofNaCl to BaS(34. 
rJaoU^ Zoo. 44 

From an inspection of these ratios it will be noted that 
of the twelve, each has its reciprocal; ia)'\ :^ — ^^ having 

Its reciprocal m (d) ', ^^^, ; (6) , ^^^^ m (^) , 



BaCl2 ' '^ ' BaS04 '^' ' BaCl2 ^ 
and so on throughout. This then eliminates six of the 
twelve as superfluous, for knowing the value of the factor, 
the value of its reciprocal follows. To multiply by a 
number is equivalent to dividing by its reciprocal and 
vice versa. Consider the reciprocal factors (b)" and {g)". 
Problems involving these may be typified as: 

Q))" How many grams of barium chloride are required 
to produce X grams of barium sulphate? By employing 
factor Q))", the answer is obviously 0.8923 X g. 

{g)" How many grams of barium sulphate will be pre- 
cipitated by X grams of barium chloride? Factor {g)'' 
gives the answer, 1.1207 X g. Supposing factor for (6)" 
only were known and problem (^)''is to be solved. This 
is done by simply dividing by the reciprocal factor for 
i}))"j the result being, 

^^= 1.1207 Xg., as before.i 

^ When a factor is given and its reciprocal is desired, instead of 
dividing by the factor given, if logarithms are used and the loga- 
rithm of the factor is known, it is more convenient to merely add the 
cologarithm. 



RATIOS 11 

Factors as given in chemists' handbooks are usually 
tabulated in four columns: Given, Sought, Factor and 
Logarithm of the factor. In problem (h)" barium sul- 
phate is given; barium chloride is sought and the factor 
is 0.8923. In problem (g)" these conditions are reversed; 
barium chloride is given and barium sulphate is sought; 
consequently the operation is reversed.^ 

Percentage Composition of Chemical Compounds. — 
The law of constancy of composition, stating that every 
well-defined chemical compound always consists of the 
same elements united in invariable proportion by weight, 
furnishes a means of calculating the ratio composition of 
any compound and consequently the weight of any ele- 
ment or group of elements contained in any weight of 
that compound. The compound MgS04 • 7 H2O is com- 
posed of 

Mg 24 32 

MgSO. • 7 H.0 = 246:50 = ^'^^^ P^""*^ °^ 

magnesium, per unit weight MgS04 • 7 H2O, 

S 32 07 

MgSa-7H.O = 246:50 = "-^^"^ P^""*^ °^ 

sulphur per unit weight MgS04 • 7 H2O. 

110 11(16) ^_,,„ , , 

MgS04.7H2Q =246:50 = ^'^^^^ ^^''' "^ 

oxygen per unit weight MgS04 • 7 II2O, 

^ See Chem. Ann., p. 13. In this tabulation, both factors are 
given, one being the reciprocal of the other. In the problem (6), 
barium chloride is "required," while barium sulphate is "found." 
An inspection of the arrangement of the tables will show when the 
column headed "a" is to be used, as the line giving the conditions 
of "required" and "found" are given on the line marked "^." 
The factor (g)'^ is found in column "5" as the line marked "5" 
fuljfins the conditions of the problem. 



12 CHEMICAL CALCULATIONS 

14 H 14(1.008) 



MgS04 • 7 H2O 246.50 



= 0.0572 parts of 



1.0000 
hydrogen per unit weight MgS04 • 7 H2O. 

Or, if magnesium sulphate is regarded as made up of the 
radicals, MgO • SO3 • 7 H2O, then 

MgO 40.32 



MgS04 • 7 H2O 246.50 

SO3 ^ 80.07 

MgS04 • 7 H2O 246.50 

7H2O _ 7 (18.016) 

MgS04 • 7 H2O 246.50 



= 0.1636 parts of 

magnesium oxide, 

= 0.3248 parts of 

sulphur trioxide, 

= 0.5116 parts of water. 
1.0000 



Equivalents. — If a compound contains the elements 
of a radical, even though it be impracticable actually to 
convert the compound into the desired radical, yet the 
amount of the radical which might theoretically be derived 
if the conversion were possible is readily calculated, though 
the equations may be lacking. Thus : what is the arsenic 
trioxide equivalent of potassium arsenite? Representing 
the equivalence by the sign =, 

KsAsOs — > AS2O3. 
But arsenic trioxide contains two atoms of arsenic in the 
molecule while potassium arsenite contains only one atom 
of arsenic; hence two molecules of potassium arsenite 
must be taken, 

2 K3ASO3 = AS2O3 
and the factor is 

AS2O3 197.92 



2 K3ASO3 2 (240.22) 



0.4112. 



RATIOS 13 

Successive Reactions. — When a compound is con- 
verted into some other substance by successive steps, and 
the amount of one of the initial compounds, needed to 
furnish a certain amount of a final product, is required, 
it is unnecessary to calculate the intermediary products. 
For example, the Solvay process for the manufacture of 
sodium carbonate depends upon the equations: 

NH3 + H2O + CO2 = NH4HCO3, 

NaCl + NH4HCO3 = NaHCOs + NH4CI, 

2 NaHCOa = NaaCOa + H2O + CO2. 

It is desired to know the amount of ammonia involved in 
the production of one ton of sodium carbonate. It is 
needless to calculate the amount of acid ammonium car- 
bonate and of acid sodium carbonate; it is necessary only 
to notice that one molecule of ammonia yields one mole- 
cule of acid ammonium carbonate which in turn yields 
one molecule of acid carbonate, while it requires two mole- 
cules of acid sodium carbonate to yield one molecule of 
sodium carbonate. The equations must then be modified 
to 

2 NH3 + 2 H2O + 2 CO2 - 2 NH4HCO3, 
2 NaCl + 2 NH4HCO3 = 2 NaHCOs + 2 NH4CI, 
2 NaHCOs = Na2C03 + H2O + CO2. 

The sodium carbonate depends upon acid sodium carbon- 
ate for its production, the weight of which, requisite to 
produce one ton of sodium carbonate, being 

^N^CO?' ^^=^ *^^^ ^^ NaHCOs; (1) 

and X tons of acid sodium carbonate require 

™HCq' X ^ = ^ *«^^ ^^ NH4HCO3; (2) 



14 CHEMICAL CALCULATIONS 

and the production of Y tons of ammonium acid car- 
bonate call for 

^ ^ X Y = Z tons of ammonia. (3) 

Substituting in (2) the value of X found in (1), the re- 
sult is 

2NH4HCO3 2NaHC03 y -, _ y 
2 NaHCOs NaaCOs ^ ' 

and substituting in (3) this value of Y and canceling; 
2NH3 ..^^^^i3iiGQ^..-2^?sS€03 



x. v:r:::rr- x ^vr^r- x i =z. 



Then the number of tons of ammonia sought is 
2NH3 

Na2C03^ ' 

therefore 

^ = n^l^ X 1 = 0.3213 ton. 
106.00 

It is not necessary to go through all this; the above is 
offered in proof. These reactions are summed up as 
follows : 

2 NH3 ^ 2 NH4HCO3 = 2 NaHC03 = NaaCOs; 

therefore 

2 NH3 = NaaCOs, 

and the ratio is determined.^ 

^ Another method of arriving at the same result may be exem- 
plified by taking the three equations given above, canceling like 
molecules on opposite sides of the equality sign and adding: 
1 1 

2NH3+^,H20+\^C02 =-^^?H5»GOs- 
2 NaCl H-^ ^^IVICO, =-^^^aHGO^ + 2 NH4CI 

- ^-NaSCQ^ = Na2C03 + "S^ +-€ Q^ 

2 NH3 + 2 NaCl + H2O + CO2 = Na2C03 + 2 NH4CI 
from which it is at once evident that 2 NH3 = Na2C03. 



RATIOS 15 

Factors Other than Chemical. — Factors are of very 
general application, tables of lengths, weights and other 
measures coming under this head. Thus, three feet make 
one yard. The factors are f = 3 for the conversion of 
yards into feet and its reciprocal ^ = 0.333 is the factor 
to convert feet into yards. Such examples might be in- 
stanced indefinitely. Take the calculation of factors for 
the conversion of the metric system of lengths into the 
English system.^ Having given 



Metric system. 



10 decimeters = 1 meter 
100 centimeters = 1 meter 
1000 millimeters = 1 meter 
1000 meters = 1 kilometer 



English system. 



12 inches = 1 foot 

3 feet = 1 yard 

5280 feet = 1 mile 



and also the fundamental relation, 1 meter = 39.37 inches. 
The factor for the conversion of meters into feet is 

^ = 3.2808, 

which means that 1 meter = 3.2808 feet. Its reciprocal, 

^^ 0.30480, or — ^ = 0.30480, 



39.37 — — ' - 3.2808 

is the factor for the conversion of feet into meters. In the 
same way, 

39 37 

' = 1.0936: factor of meters to yards. 
oo 

36 1 

W^ , or = 0.91440: factor of yards to meters. 

then as 1 inch is yV^h of a foot and 1 foot = 0.30480 meters, 

-^-T^ — = 0.02540: factor of inches to meters, 

1 See Chem. Ann., pp. 469-480. 



16 



CHEMICAL CALCULATIONS 



and as 1 meter = 100 centimeters, then 1 inch 
centimeters. So a table may be constructed: 



2.540 



Given. 


Sought. 


Factor 
(or number). 


Logarithm. 


meters 

yards 

meters 

feet_ 

centimeters 

inches 


yards 

meters 

feet 

meters 

inches 

centimeters 


1.0936 
0.9144 
3.2808 
0.3048 
0.3937 
2.5400 


0.03886 
1.96114 
0.51598 
1.48402 
1.59517 
0.40483 



PROBLEMS 



I. Given the reaction: 



PbCl2 + K2Cr04 
278.02 



PbCr04 + 2KCl. 
323.2 



(a) What is the factor of lead chloride to lead chromate? 
{b) If 0.1784 g. of lead chromate are precipitated by an excess 
of potassium chromate from a solution containing lead chloride, 
how many grams of lead chloride were present? (c) How many 
grams of lead chromate are obtained from 1.000 g. of lead 



chloride? (d) 
chromate? 



How many grams of lead in 0.7325 g. of lead 



Ans. (a) 



PbCls 



278.02 



PbCr04 323.1 
(6) 0.1784 X 0.8605 



= 0.8605. 



(c) 



id) 



1 



0.8605 
Pb 



0.1535 g. PbCla. 

= 1.1622 g. PbCr04. 

207 1 

^ X 0.7325 = 0.4695 g. Pb. 



PbCr04 323.1 
II. Hydrous sodium carbonate may be converted into the 
anhydrous salt by heat according to the equation: 

NasCOs • 10 H2O = NaaCOs + 10 H2O. 
286.16 106.00 180.16 

(a) How many pounds of anhydrous sodium carbonate may 
be obtained from 15.000 pounds of the crystalHzed salt? {b) 



RATIOS 



17 



What is the factor of hydrous sodium carbonate to anhydrous 
sodium carbonate? (c) If 17.000 pounds of hydrous sodium 
carbonate are converted into the anhydrous form, what is the 
loss in weight? 
Ans. 

NaoCOs 106.00 



(a) 



(b) 



NazCOs • 10 H2O 
NaaCOa-lOHsO 



286.16 
286.16 



X 15.000 = 5.5214 lbs. NaaCOs. 



2.6996. 



NasCOs 106.00 

(c) The loss in weight is the water driven off. This problem 
may be solved in two ways: By using the factor found in (6), 
or by calculating the water directly. 

i^OOO ^ 6 2973 lbs. NasCOs remaining. 

17.000 - 6.2973 = 10.703 lbs. water driven off. 
IOH2O 180.16 



^^ Na2C03 
driven off. 



IOH2O 286.16 



X 17.000 = 10.703 lbs. water 



1. (a) What is the factor for the conversion Mg2P207 -> P2O5? 
(b) How many grams of phosphoric anhydride are contained in 
0.7256 g. of magnesium pyrophosphate? (c) What is the factor 
for the conversion of (NH4)3 PO4 • 12 M0O3 to P2O5? (d) How 
many grams of phosphoric anhydride are equivalent to 0.1500 g. 
of ammonium phosphomolybdate? Ans. (a) 0.63793. 



2. Verify the following factors: 



(&) 0.4629 g. 

(c) 0.03784. 

(d) 0.005677 g. 



Given. 


Sought. 


Factor (number). 


KsPtCle 

PbMo04 
NO 

PbS04 

Mn2P207 


Pt 

M0O3 

NaNOa 

Pb 

MnO 


0.40151 

0.39226 
2.8327 
0.68311 
0.49961 



18 CHEMICAL CALCULATIONS 

3. How many grams of chromic sulphide will be formed from 
0.7182 g. of chromic oxide according to the equation: 

2 CrsOs + 3 CS2 = 2 Cr2S3 + 3 CO2? 

Ans. 0.9460 g. 

4. How many grams of tin must be treated with nitric acid 
to obtain 2 kilos of stannic oxide? 

Sn-^Sn02. 

Ans. 1576 g. 

5. How much charcoal is required to reduce 1.500 g. of 
arsenic trioxide according to the equation: 

AS2O3 + 3C = 3 CO + 2 As? 

A71S. 0.2728 g. 

6. Iodine is liberated according to the equation: 

2 KI + Clo = 2 KCl + I2. 

(a) How many grams of iodine are liberated from 0.1726 g. of 
potassium iodide? (6) If 1.5370 g. of iodine are liberated, how 
many grams of potassium iodide were decomposed? 

Ans. (a) 0.1319 g. 
(b) 2.0105 g. 

7. Oxygen is prepared from potassium chlorate according to 
the equation: 

2KC103 = 2KCl + 3 02. 

(a) What is the yield of oxygen when 7.0680 g. of potassium 
chlorate are decomposed? (6) How many grams of potassium 
chlorate must be decomposed to get 2.0000 g. of oxygen? 

Ans. (a) 2.7681 g. 
(6) 5.1067 g. 

8. In the compound CaCOs • 3 Caa (P04)2, {a) how much 
phosphorus is contained in 5.000 g.? {b) How much phosphoric 
anhydride in the same amount? Ans. (a) 0.9031 g. 

(6) 2.067 g. 

9. Sulphur dioxide may be produced by the reaction: 

Cu + 2 H0SO4 = CUSO4 + 2 H2O + SO2. 



RATIOS 19 

(a) How much copper and (6) how much of a 93.20% H2SO4 
must be taken to obtain 64.00 g. of sulphur dioxide? 

Ans. (a) 63.50 g. 
(6) 210.3 g. 

10. How much superphosphate can be made from one ton of 
calcium phosphate, 93.50% pure? The reaction is 

Ca3 (P04)2 + 2 H2SO4 = 2 CaS04 + CaH4(P04)2. 

Ans. 0.7056 ton. 

11. Sulphuric acid is made according to the equation: 

2 S + 3 O2 + 2 H2O = 2 H2SO4. 

(a) If brimstone containing 97.00% sulphur is used, how much 
sulphuric acid is obtained from one ton? {h) If pyrites con- 
taining 96.00% FeS2 is used to furnish the sulphur, how many 
tons are required to yield a ton of sulphuric acid? 

Ans. (a) 2.9667 tons. 
(6) 0.6371 ton. 

12. Sulphuric acid reacts with sodium hydroxide thus: 

H2SO4 + 2 NaOH = Na2S04 + 2 H2O. 

If 0.2073 g. of sulphuric acid are added to 0.1705 g. of sodium 
hydroxide, (a) how much sodium sulphate is formed and {b) 
which is left over, caustic or acid, and how much? 

Ans. 0.3003 g. 

0.0014 g. NaOH. 

13. (a) What is the percentage of manganese in pure potassium 
permanganate (KMn04)? (b) If contaminated to an extent of 
2.00% of impurities? Ans. (a) 34.76%. 

(6) 34.( 



14. Dolomite is a double carbonate of calcium and magne- 
sium which may in some cases be represented by CaCOs • MgCOs. 
(a) What are the percentages of magnesium carbonate and of 
calcium carbonate in such a sample? (6) How many pounds 
of magnesium carbonate and calcium carbonate in a ton of this 



20 CHEMICAL CALCULATIONS 

dolomite? (c) How many tons of Epsom salts, MgS04 • 7 H2O, 
can be obtained from a ton of dolomite? 

Ans. (a) 54.27% CaCOg, 
45.73% MgCOs. 
(6) 1085.4 lbs. CaCOa, 
914.6 lbs. MgCOa. 
(c) 1.337 tons. 

15. What are the percentages of the elements in ammonium 
phosphomolybdate if it is (NH4)3P04 • 12 M0O3 • 3 H2O? 

Ans. N = 2.18%), = 35.63%, 
H = 0.93%, Mo = 59.65%, 
P = 1.61%. 

16. Regarding ammonium phosphomolybdate as made up of 
the radicals NH3, H2O, P2O5 and M0O3, what is the percentage 
composition of these radicals in the molecule? See Prob. 15. 

Ans. P2O5 = 3.69%, 

H20= 4.20%, 

NH3= 2.65%, 

M0O3 = 89.47%. 

17. Potassium antimonyl tartrate (tartar emetic) corresponds 
to the formula K2H2 (C4H406)2 • Sb203. (a) What are the per- 
centages of the different elements in this compound? (b) What 
is the percentage of Sb203? (c) How many grams of tartar 
emetic must be taken to obtain 5.0000 g. of antimony? 

Ans. (a) K = 11.76%, H = 1.52%, 
C = 14.44%, O = 36.11%, 
Sb = 36.17%. 
(6) 43.39%. 
(c) 13.8245 g. 

18. 5.000 g. of arsenic (the metal) are oxidized and the oxide 
dissolved in caustic potash. How many grams of potassium 
arsenite are formed? The reactions are 

4As + 3 02 = 2As203, 
AS2O3 + 6 KOH = 2 K3ASO3 + 3 H2O. 

Ans. 16.026 g. 



RATIOS 21 

19. (a) How many pounds of salt are required to make 1500 
lbs. of salt cake (Na2S04)? (b) How many pounds of Glauber's 
salt (Na2S04 • 10 H2O) will this amount of salt cake make? 
Reactions: 

2 NaCl + H2SO4 = 2 HCl + Na2S04, 
Na2S04 + 10 H2O = Na2S04 • 10 H2O. 

Ans. (a) 1234.4 lbs. 
(6) 3402.2 lbs. 

20. How many grams of ammonium dichromate may be pre- 
pared from 500 g. of potassium dichromate according to the 
reactions: 

K2Cr207 + H2SO4 = K2SO4 + H2O + 2 CrOs, 
2 CrOa + 2 NH3 + H2O = (NH4)2 feOr? 

Ans. 428.4 g. 

21. Iodine may be obtained from potassium iodide according 
to the equations: 

NaCl + H2SO4 = NaHS04 + HCl 
4 HCl + MnOa = MnCl2 + 2 H2O + CI2 
Cl2 + 2KI = 2KCH-l2. 

How much sulphuric acid must be taken to produce 5.000 g. of 
iodine? Ans. 7.728 g. 

22. The Leblanc process for the manufacture of sodium car- 
bonate is 

2 NaCl + H2SO4 = Na2S04 + 2 HCl, 

Na2S04 + 2 C = NasS + 2 CO2, 

Na2S + CaCOs = Na2C03 + CaS. 

How many tons of sodium carbonate may be obtained from a 
ton of salt? Ans. 0.9066 ton. 

23. From the equations: 

AICI3 + 3 NH4C2H3O2 = Al (C2H302)3 + 3 NH4CI, 

Al (C2H302)3 + H2O = Al (OH) (C2H302)2 + HC2H3O2, 

2 Al (OH) (C2H302)2 + 8 O2 = AI2O3 + 7 H2O + 8 CO2. 

(a) How many grams of aluminum chloride are required to 
yield 0.3000 g. of aluminum oxide? (6) How many grams of 



22 CHEMICAL CALCULATIONS 

aluminum oxide are obtained from 0.8300 g. of aluminum 
chloride? Ans. (a) 0.7836 g. 

(6) 0.3177 g. . 

24. Chrome iron ore is Cr203 • FeO, and may be converted 
into potassium dichromate as follows : 

4 FeO . Cr203 + 4 K2CO3 + 4 CaO + 7 O2 
= 4 K2Cr04 + 4 CaCr04 + 2 Fe203 + 4 CO2. 

The calcium chromate is converted into potassium chromate, 

CaCr04 + K2SO4 = CaS04 + K2Cr04; 

and potassium dichromate is obtained from potassium chromate, 

2 K2Cr04 + H2SO4 = K2SO4 + H2O + KsCraOy. 

How many tons of potassium dichromate can be obtained from 
a ton of chrome iron ore, if the conversion is complete and the 
ore is 92.00% FeO • Cr203? Ans. 1.2092 tons. 

25. The barometer reads 30.00". What is the reading in the 
metric system?^ Ans. 76.20 cm. 

26. A piece of aluminum wire 200 mm. long weighs 0.1327 g. 
What length should be taken to make a centigram rider? 

Ans. 15.07 mm. 

27. It is required to find the height of a can to hold one quart, 
the diameter of which must be 4.5", one half inch being allowed 
at the top for air space. Ans. 4.13". 

28. In estimating the capacity of paint cans, they were filled 
to within one half inch of the top with water at 60° F. The 
weights of water were 

A, 4.169 lbs. C, 2.084 lbs. 

B, 1.042 lbs. D, 8.338 lbs. 

What were the capacities of these cans to within one-half inch of 
the top? (1 gallon = 231 cu. in.) Ans. A, 2 qts. 

B, 1 pt. 

C, 1 qt. 

D, 1 gal. 

1 The sign (') is used as an abbreviation for feet, (") for inches. 



RATIOS 



23 



29. It is desired to make a 50-cc. burette, graduated to tenths 
of a cubic centimeter, the graduations to be 2 mm. apart. What 
should be the diameter of the glass tube? Ans. 0.798 cm. 

30. Eimer & Amend's catalogue gives the following data 
about platinum foil: ''Platinum foil, medium, to^o" thick, one 
gram per square inch." Assuming the price of platinum to be 
$1.60 per gram, what would a cone for electrolysis cost, having 
a slant height of 4", diameter 3'7 Ans. $30.16. 

31. The Westinghouse handbook gives the following data 
relative to copper wire. 



Gage. 


Diam. (") 


Area sq. in. 


Lbs. per 1000 ft. 


Ft. per lb. 


10 


0.10381 


0.0081532 


31.37 


31.38 



If one foot = 30.480 cm. and one pound = 453.59 g.: (a) 
What length in centimeters must be taken to weigh 1.500 g.? 
(6) What is the weight in pounds of 7.000 m.? 

Ans. (a) 3.213 cm. 
(6) 0.73191b. 

32. Working up from the datum, one inch = 2.540 cm., verify 
the following : 



Given. 


Sought. 


Factor. 


square centimeters 


square inches 


0.1550 


square inches 


square centimeters 


6.4516 


cubic centimeters 


cubic inches 


0.06102 


cubic inches 


cubic centimeters 


16.387 


square meters 


square yards 


1 . 1960 


square yards 


square meters 


0.8361 


cubic meters 


cubic yards 


1.3079 


cubic yards 


cubic meters 


0.7645 



33. The internal diameter of a spherical glass bulb is 4.382' 
what are its contents in cubic cm.? Ans. 721.9 cc. 



24 CHEMICAL CALCULATIONS 

34. A Dumas flask for the determination of molecular weights 
measures 8.5" in inner diameter. Neglecting the volume of the 
drawn-out neck, what is its capacity in liters? Ans. 5.27 L. 

35. Oil of vitriol is shipped in iron containers. A scow is to 
be fitted with two tanks running fore and aft. It is able to carry 
200 tons of the acid. If the tanks are to be 60' long, what 
must be the diameter of the tanks to carry this acid? One 
cubic foot of oil of vitriol weighs 114.47 lbs. Ans. 6.09'. 

36. (a) How many liters in a cubic foot? (6) How many 
liters in a gallon? (c) How many quarts in two liters? 

Ans. (a) 28.317 L. 

(b) 3.785 L. 

(c) 2.113 qts. 

37. A circular piece of filter paper 15 cm. in diameter yields 
0.00025 g. of ash. (a) What will the weight of ash be from a 
circular piece 5.5 cm. in diameter? (b) From a piece of the 
same kind of paper 3" square? Ans. {a) 0.00003 g. 

(6) 0.00008 g. 

38. A tank measures 15.00' X 20.00' X 6.00'. How many 
liters will it hold? Ans. 50,969 L. 

39. From the fundamental relation between meters and 
inches, (a) derive the ratio of square feet to square meters, 
(b) square meters to square yards, (c) If a wall has an area of 
2057 square feet, how many square meters does it contain? 
(d) If cloth costs $2.50 per square meter, what will it cost per 
square yard? Ans. (a) 0.09290. 

(6) 1.1960. 

(c) 191.1 sq. m. 

(d) $2.09. 

40. (a) How many cubic inches in a liter? (6) How many 
cubic meters in an excavation 72' X 50' X 12'? 

Ans. {a) 61.025 cu. in. 
{b) 1223.2 cu. m. 

41. Derive the value of a kilometer expressed in miles from 
the fundamental unit; one meter = 39.37 inches. 

Ans. 1 km. = 0.62137 mi. 



RATIOS 25 

42. A nautical mile is the length of one minute of the earth's 
circumference at the equator and is equal to 1.1527 statute or 
land miles, (a) What is the circumference of the earth at the 
equator? (6) What is the diameter of the earth at the equator 
in meters? (c) In kilometers? (d) How many feet in a nau- 
tical mHe (knot)? Ans. (a) 24,898 mi. 

(b) 12,755,000 m. 

(c) 12,755 km. 
id) 6086 ft. 



CHAPTER II 
APPROXIMATE NUMBERS 

An abstract number is accurate as stated, but when 
a number obtained by measurement represents some 
multiple of units obtained by measurement or manipula- 
tion, it is not absolutely accurate, but only relatively 
so. This error is called the experimental error or the 
error of measurement. For example, suppose a sub- 
stance is weighed on a balance sensitive to a milligram, 
and equilibrium is established with 1.628 g. as a counter- 
balance. Should this same weight be determined by a 
balance sensitive to tenths of a milligram, it cannot be 
predicted that it will weigh 1.6280 g.; in fact, it may 
weigh anything between 1.6275 and 1.6285 g. These 
figures are the extreme limits of variance. The error in 
weighing on the balance sensitive to a milligram may be 
less than this; it can only be said with certainty that it 
is not more. There must be a figure beyond which there 
is uncertainty. Such numbers are approximate numbers. 
By convention, when a number expresses a multiple of 
some unit which has been experimentally determined, or 
derived, it is customary that the last figure given is 
the last figure known with any degree of certainty. In 
the example above, the last figure known is 8 in the 
thousandths place. The uncertainty is dz5 in the ten 
thousandths place or ±0.0005 g. This is the maximum 
apparent error. 

A number expressing some multiple of a unit, experi- 
mentally determined, or derived, always contains some 

26 



APPROXIMATE NUMBERS 27 

error, which is great or small, according to the difficulty, 
or the delicacy of the operation by which it is determined; 
hence there must be some figure in the result beyond 
which there is uncertainty. In atomic weight tables the 
element iodine will be found stated to five significant 
figures, while the majority of the atomic weights are given 
only to three or four significant figures. This is for the 
reason that this element, a very important one in analyti- 
cal chemistry, has been very carefully studied, and also 
because of some of its properties, by taking advantage of 
which it (the atomic weight) may be determined with the 
degree of accuracy indicated. 

This is the convention, and strictly speaking is seldom 
accurately followed, as frequently results having an error 
in the third place are reported as far as the fourth. 

There are methods of abbreviated multiplication and 
division, which eliminate superfluous figures, and indicate 
the extent to which the result may be relied upon in the 
use of approximate numbers. Logarithms, however, do 
all this with the same degree of certainty, and with much 
less mental fatigue, especially when many extended oper- 
ations are to be performed. Logarithms themselves are 
only approximate numbers, and when employed should 
have a degree of accuracy in their last significant figure 
equal to, or in excess of, the accuracy of the last significant 
figure in the approximate numbers involved. For this 
reason, four- and five-place tables are generally employed, 
with the balance of ^ opinion in favor of the five-place 
tables for the usual calculations performed by the chemist. 
In chemists' handbooks the logarithms of analytical 
and other factors are given to five places in the mantissa, 
though many of the figures are reliable only to four places, 
yet quite as many are accurate to five significant figures. 
Also, to obtain four significant figures from four-place 



28 CHEMICAL CALCULATIONS 

tables requires interpolation, while the five-place tables 

give four figures by direct reading and five by interpola- 

tion.i 

Addition and Subtraction of Approximate Numbers. 

— It is required to add the following atomic weights, 

assuming the maximum apparent errors, in order to 

obtain the molecular weight of aluminum hydroxide, 

A1(0H)3. 

Al =27.1 

30 = 3X 16.00 =48.00 

3H = 3X 1.008 = 3.024 

A1(0H)3 =78.124 

The last figure (4) in the thousandths place is manifestly 
uncertain, for the reason that 27.1 has an error of d=5 
in the hundredths place. Strictly speaking, the sum is 
accurate only to the extent of the tens place, and the 
result would read 78.1, but the underlined figure is re- 
tained for the reason that less error is caused by its reten- 
tion than by its rejection.^ 

* Logarithms have been employed in the solution of problems in 
this book, except where the operation was very simple. In such 
cases the 10" Mannheim sHde rule was used. The author would 
strongly recommend this simple device for the checking of results 
in the laboratory. It cannot be used exclusively, as the accuracy 
of the rule is only to three figures in the higher numbers. The 
mantissa only of a logarithm may be used, neglecting the charac- 
teristic, the position of the decimal point being determined by in- 
spection. A chemists' slide rule, manufactured by Kruffel & Esser 
and devised by the author is now on the market. It is designed to 
facilitate the computation of such problems as are found in this book 
and is recommended to students, as by its use much time is saved in 
calculating the answers to the problems. 

2 This convention has not been strictly followed in the tabula- 
tion of molecular weights. See Chem. Ann., pp. 39-47; 100-336. 
One more figure is carried when the number is to be used in further 
computation. 



APPROXIMATE NUMBERS 29 

Subtraction of approximate numbers leads to the same 
results. 

Multiplication of Approximate Numbers. — Multiply- 
ing the number 1.862, containing maximum apparent 
error, by 0.6257, the results obtained are: 

(a) By straight multiplication, assuming no errors, 
1.862 X 0.6257 = 1.1650534. 

(6) By assuming the maximum apparent error acting 
in a negative (subtractive) direction, 

1.8615 X 0.6257 - 1.16474055. 

(c) By assuming the maximum apparent error acting 
in a positive (additive) direction, 

1.8625 X 0.6257 = 1.16536625. 

In practice, having a problem of this nature, it is not 
known whether the error is working in a negative direc- 
tion (decreasing) as in (b) or positive direction (increas- 
ing) as in (c) . If there is any error at all it must be either 
positive or negative and it is not known which, nor is it 
known to what extent; it is only known that it is not 
more than d=5 in the ten thousandths place; consequently, 
the most probable result is obtained from (a) and is 
written 1.165, the last figure being uncertain as is seen by 
an inspection of the results of (6) and (c) . 

Again, multiplying 57.2, containing maximum apparent 
error, by 15.27: 

(a) Assuming no error, 

57.2 X 15.27 = 873.444. 

(h) Assuming the maximum apparent error as being 
negative, 

57.15 X 15.27 = 872.6805. 

(c) Assuming the maximum apparent error as being 
positive, 

57.25 X 15.27 = 874.2075. 



30 CHEMICAL CALCULATIONS 

The result as obtained from (a) is 873. It is evident that 
when a result is obtained by a multiplication involving a 
unit multiple (approximate number) given to the degree 
of accuracy adopted by convention, the result may be 
depended upon only to the same number of significant 
figures as is contained in the approximate number. It 
often happens that both the numbers to be multiplied 
are approximate numbers. Take the case of multiplying 
572 and 5725 each containing maximum apparent errors. 
The result is 3274700 and is reliable only the third figure, 
for the reason that 572 contains only three significant 
figures, hence the product is accurate to the same number 
of places only. 

Division of Approximate Numbers. — Take a case of 
division involving approximate numbers; 5724 contains 
maximum apparent error, 172 is absolute, 

^794 1 72 

(a) ^^ = 33.2791 • • ^ : ^ = 0.0300489 . . . 

(6) "^ = 33.2762 ...■■^,= 0.0300515 . . . 

^^794 ^ 172 

(c) -^ = 33.2821 . . . .-57^= 0.0300462 . . . 

Again, if the number which has a maximum apparent 
error contains only three significant figures, as 152, 
while 1645 is absolute (contains no error), the following 
results are obtained: 

(^0 T^ = 10.8223 . . . : .^ = 0.0924012 . . . 
^ 152 1645 

(^) So = ''■'''' • • • ^ 1645 = 009^09^2 . . . 



APPROXIMATE NUMBERS 31 

The limit of accuracy, evidentlyj is the same as in the mul- 
tipHcation of approximate numbers. 

Recapitulation. — I. The sum obtained by adding 
approximate numbers is retained to one place beyond the 
indeterminate figure.^ 

II. Subtraction follows the same rule as addition. 

III. The result obtained by multiplication involving 
approximate numbers is retained to the same number of 
significant figures as is contained in the approximate 
number having the smallest number of significant figures. 

IV. Division follows the same rule as multiplication. 
It does not come within the province of this book to 

give instruction in the use of logarithms, but a few re- 
marks advocating their use are appropriate. A consider- 
ation of approximate numbers shows that results obtained 
in the manipulation of such numbers are to be depended 
upon to a degree commensurate with the degree of accu- 
racy of the approximate numbers employed. In most 
chemical operations, readings are taken to four or five 
significant figures. Thus a burette is read to four places 
with the last figure estimated. The average masses 
determined by the analytical balance seldom exceed six 
figures, and allowing for tare, the net weight seldom ex- 
ceeds five, and is generally accurate only to four figures. 
These approximate numbers, entering as factors into 
subsequent calculations, carry with them their inherent 
errors, with the consequence that most results are stated 
to four significant figures only. In an analysis the con- 
stituents are generally given to hundredths of a per cent. 

1 One more figure is carried when the result is to be used in 
further computation; this applies also for subtraction, multiplication 
and division. 



CHAPTER III 
INTERPOLATION 

Functions. — When two quantities are interdepend- 
ent, one is said to be a function of the other. The 
relation may be very simple, as between pounds and kilo- 
grams. An increase of weight in a body of x pounds 
will cause an increase of y kilograms, x being a function 
of y. Expressed mathematically, it is a; = ky. Conse- 
quently, knowing the ratio of x to y, x being given, y is 
readily determined. Let x be the number of pounds and y 
the number of kilograms change in weight of a system; to 
convert x pounds into kilograms it is only necessary to 
multiply by the factor k. In this instance, k = 0.45359; 
so the expression becomes x = 0.45359 y. This is direct 
variation, and by plotting the curve, measuring one of the 
units along the ordinate and the other along the abscissa, 
a straight line is obtained. Such curves are seldom 
plotted.^ 

Graphic Representation of Functions. — Any tabu- 
lation, when plotted, yields a curve, the simplicity of 
which depends upon the particular case under consider- 
ation. When two quantities vary directly the curve is 
a straight line; ^ when varying inversely, an hyperbola 

^ A tabulation of results obtained by solving the different values 
of one quantity in terms of the other, where the resulting curve is a 
straight line, is sometimes contained in handbooks for quick reference. 

2 When the equation is of the first degree, that is, when it does 
not contain either of the quantities raised to any power other than 
the first. 

32 



INTERPOLATION 



33 



10 
9 
8 

7 

6 
<D 

a 

Z5 

o 

Q. 

4 
3 
2 
1 
























( 






X 


PLATE I 
= .45559 


y 




















































' 
















/ 














/ 












/ 














/ 















KILOGRAMS 

Plate I. 



34 



CHEMICAL CALCULATIONS 



results. Any value of either x or 2/, in terms of the other, 
is as accurately or more accurately obtained by using the 
factor in the equation under consideration as by consult- 
ing the curve. 

Plate I represents the curve of pounds equivalent to 
kilograms; any value of x in terms of y will fall on the 
line OA. 



1.0 

.9 
.8 
.7 

2 .6 

H 

%■' 




-J .4 

Z 
.2 








































^^ 


^ 








PLATE 
Log.y- 


n 








^ 


















/ 


y 


















/ 




















/ 






























































1 






















1 
























1 





















4 5 6 

NUMBER 

Plate II. 



10 



Plate II represents the relation existing between a 
number and its logarithm; i.e.^ the curve fulfils the equa- 
tion log y = X. (In this case, to obtain values of y in 
terms of x involves extended mathematical calculations; 



INTERPOLATION 



35 



consequently these values are determined and once for all 
tabulated.) Any value of x in terms of y, or vice versa, 
will fall on the curve. 



90 



70 



^60 

ID 



40 



30 



20 



P/ate V shows area 
wifhin dotted lines 
enlarged. 



PLATE m 
Show in a the relation 



between the percent 
composition and the 
Beaume of sulphuric acid. 





PI a te 17 shows area 
within dotted lines 
■enlarged- 



20 30 40 50, 

DEGREES BEAUME 

Plate III. 



60 



70 



Change of Percentage Composition with Baume for 
Sulphuric Acid. — Plate III shows the relation between 
the percentage composition and the strength in degrees 



36 CHEMICAL CALCULATIONS 

Baume of sulphuric acid.^ The ''Baume" will be, for 
the present, used as the weight of a unit volume.^ An 
increase in the Baume of sulphuric acid is accompanied 
by an increase in the percentage of sulphuric acid (H2SO4) ; 
so if X represents the percentage of sulphuric acid (H2SO4), 
in a given sample, and y its Baume, the relation be- 
tween these quantities is represented by the equation 
X = {f)y.^ 

The distances along the axis OY represent the degrees 
Baume (° Be.) of sulphuric acid and those along OX the 
percentage composition; if one varied directly as the other 
a straight line would result. Let 5° Be. correspond to 
5.28%H2S04 (point E), then 60° Be. would correspond 
to -65O- X 5.28 = 63.36% H2SO4 (point F'). Connecting 
these points the curve OA is obtained. 

OB is the true relationship between degrees Baume 
and the percentage sulphuric acid, determined experi- 
mentally; the curve continually diverging from the 
straight line OA. Owing to this divergence, acid of 
60° Be. contains 77.67% H2SO4 (point F) instead of 
63.36%, the percentage calculated on the assumption 
that the per cent H2SO4 varied directly as the Baume; 
i.e., that the curve was the straight line OA. This differ- 
ence, 14.31%, is represented by the distance FF\ The 
divergence continually increases as the acid approaches 
100% H2SO4; in fact, as the strength of the acid ap- 
proaches and passes 93.19% H2SO4, corresponding to 
66° Be., a very slight change in the Baume strength 

1 The data has been taken from the tables of Ferguson and 
Talbot for Sulphuric Acid. See Chem. Ann., pp. 388-393. 

2 The method of determining the Baume of a liquid will be dis- 
cussed at greater length under Specific Gravity, p. 65. 

3 The value of the functions {f)y in the equation x = {f)y in this 
case varies with the concentration. The value of x corresponding 
to y must be determined experimentally. 



INTERPOLATION 37 

causes a considerable change in the per cent H2SO4, the 
curve approaching perpendicularity. It is for this reason 
that the method is not employed for acids of a Baume 
strength greater than 66"^ Be. 

Interpolation. — Consider the curve OA included be- 
tween the points C and D'; being given: 

C = 48.00° Be. corresponding to 50.69% H2SO4 

D' = 52.00° Be. corresponding to 54.91% H2SO4 

difference = 4.00° Be. 4.22% H2SO4 

It is required to find the per cent H2SO4 corresponding to 
50° Be. (point m') ; or, having the abscissa of the point 
m' lying on the line OA, it is required to find its 
ordinate. The abscissa distance of CD' is 4.00; the 
difference between the abscissas of C and D\ The ab- 
scissa distance of Cm' is the difference in Baume between 
C and m'; 50.00° Be. - 48.00° Be. = 2.00° Be. But the 
abscissa distance of CD' corresponds to the ordinate 
distance 4.22; hence the ordinate distance of Cm' is 

?^ X 4.22 - 2.11. The ordinate of m' is then 50.69 + 

2.11 = 52.80, the percentage of sulphuric acid correspond- 
ing to 50.00° Be., if these quantities varied according to 
the curve OA. 

This is interpolation. It can only be employed when 
the curve, between the limits given, may be considered a 
straight line. The curve OB showing the relation between 
the Baume strength and the percentage strength of sul- 
phuric acid has been determined experimentally. It is 
found to depart considerably from a straight line, but 
between two points on the curve in close enough proximity 
the line is sensibly straight. ^ This is shown in Plate IV, 

^ That is, the error is less than the errors of the other factors 
entering into consideration. 



38 



CHEMICAL CALCULATIONS 



TC 






ri 




_. J 




7 7 




/ / 


P?LA7'^ L^ 






7 y 


Lee il 6f hhrte n 


QI ■ 1 I X 




. 1 J 




■ ^ / z 




-/ ^-T-- 




2Ji/ 




k/ ?^ 




^v 5 




5Z «'^ 




.^/ 12 




^S 45 _ ± 








z z 




/^ v^v 




/ sO 




t^' L 


z 


/ 


^ 






1 ^ .f' 


7 


7 -^ 


z ^ 


i^- - /"^ 


/ / 




^ l 


_L ^5^ 


Z .^ 


" ''iCi^ 


a.ri J. Z. 






^ >' 


f &c 


-(^^ 


Z J 




/. z 


y 




/ 


7 r 


7 


I Z 


7 X 


/_ J 


/ 




7 


hF> ^ -/ 


_Z 


&5 2 2 


25' 




^ 




/ 


^ -/ 


f4»' 


12 2 




/ / ^ 




^ y ^ 




Z Z / 




.r.- 7 A' 




50 ^^ u^^ 




40 45 6 


, 55 60 



DEGREES BEAUME 

Plate IV. 

an enlargement of a small section of Plate III in the 
region of C'T>' . 

Working with the curve OB, the following data being 
given : 

C = 48.00° Be. corresponding to 59.32% H2SO4 

D = 52.00° Be. corresponding to 65.13 % H2SO4 

difference = 4.00° Be. 5.81%oH2SO. 



INTERPOLATION 39 

It is required to find the percentage of sulphuric acid 
corresponding to 50.00° Be., or the ordinate of m} 
50.00 - 48.00 = 2.00 = the abscissa distance of Cm, 
4.00 = the abscissa distance of CD, 
5.81 = the ordinate distance of CD, 
then the ordinate distance of Cm is 

consequently the ordinate of Cm, or the per cent of sul- 
phuric acid corresponding to m, is 

59.32 + 2.91 = 62.23%. 
This value, obtained by interpolation, agrees fairly closely 
with the value found experimentally, 62.18%; the dis- 
crepancy is 0.05%. 

Once more, working with the curve OB, given 

E = 5.00° Be. corresponding to 5.28% H2SO4 
F = 60.00° Be. corresponding to 77.67% H2SO4 
difference = 55.00° Be. 72.39% H2SO4 

It is required to find the per cent of sulphuric acid corre- 
sponding to 50.00° Be. 

50.00° - 5.00° = 45.00° = the excess of ° Be. of required 

acid over 5.00° B^., the 
° Be. below. 
55.00° = the interval in ° Be. between 

the points given. 
72.39% = the interval in per cent 
H2SO4 corresponding to an 

interval of 55.00° Be. 
4-^ on 

J^ X 72.39 = 59.23%) H2SO4 corresponding to an 
55.00 

increase of 45.00° Be. 
59.23+5.28 = 64.51% H2SO4 corresponding to 50.00° B^. 

1 Note that the m' and m correspond to the same Baum6. 



40 CHEMICAL CALCULATIONS 

This corresponds to point m" . The difference between 
the per cent of sulphuric acid found by this interpola- 
tion, 64.51%, and the per cent of acid found by the previ- 
ous interpolation, 62.23%, is due to the wide separation 
of the points (E) and {¥) as against the relative proximity 
of the points (C) and (D) of the previous interpolation, 
between which intervals the determined curve OB is 
almost a straight line, m" must fall on a point in a 
straight line connecting (E) and (i^). The difference, 
64.51% - 62.23% = 2.29%, is due to the curvature of 
OB between the intervals given, and is represented by 
the distance mm" . 

From these considerations, it becomes apparent that to 
interpolate, the curve must be sensibly a straight line. 
To obtain this condition, when the curve as a whole 
deviates from a straight line, the tabular difference must 
be small enough, or the fixed points near enough, that the 
curve between the points given is sensibly straight. Con- 
sequently, the greater the curvature, the smaller must be 
the interval between the given points to make the inter- 
polation accurate to the desired degree. 

Extrapolation. — Again, working with results experi- 
mentally determined (as represented by the curve 05), 
being given 

40.00° Be. corresponding to 48.10%o H2SO4 

41.00° Be. corresponding to 49.47 % H2SO4 

difference = 1.00° Be. 1.37% H2SO4 

It is required to find the percentage of sulphuric acid 
corresponding to 42.00° Be. 42.00° Be. does not lie 
between the points given, but it may be calculated if 
the curve is assumed to be sensibly straight between the 
intervals 40.00° Be. and 42.00° Be.; i.e., if within these 
hmits the Baume is proportional to the percentage of 



INTERPOLATION 41 

sulphuric acid without too much error. On this assump- 
tion the percentage of sulphuric acid corresponding to 
42.00° Be. is 

49.47 + 1.37 = 50.84% H2SO4. 

As a matter of experiment, 42.00° Be. corresponds to 
50.87% H2SO4, hence an error has been made of 0.03% 
H2SO4 (50.87 - 50.84) which is relatively small. This is 
extrapolation and is resorted to with caution for a reason 
which will appear in the following. Given 

65.75° Be. corresponding to 91.80% H2SO4 

66.00° Be. corresponding to 93.19% H2SO4 

difference = 0.25° Be. 1.39% H2SO4 

It is required to find the percentage of sulphuric acid 
corresponding to 66.25° Be. On the same assumptions 
as above, the percentage of sulphuric acid corresponding 
to 66.25° Be. is 

93.19 + 1.39 = 94.58% H2SO4. 

As a matter of experiment 66.25° Be. is found to corre- 
spond to 95.28% H2SO4, consequently an error of 0.70% 
H2SO4 (95.28 — 94.58) has been made by extrapolating 
notwithstanding the small interval taken (0.25° Be.). 
In other words, the curve is not sensibly a straight line 
between the intervals given (65.75° Be. and 66.25° B6.). 
This is shown in the large scale Plate V, page 42. 

PROBLEMS 
I. Given 

60° B6} corresponding to 77.67% H2SO4, 

61° Be corresponding to 7943% H2SO4. 

* The Baume is given in the tables to but two significant figures 
though they are accurate to two places to the right of the decimal 
point, 60° Be. being accurate to 60.00° Be. 



42 



CHEMICAL CALCULATIONS 





















Deh 


PLATE V 
111 of plate 


m 






































95.28 












9 94.58 

1 
ji 












1 













































'64 65 66 67 68 , 

DEGREES BEAUME 

Plate V. 



70 



INTERPOLATION 43 

(a) If the Baume of a sample of sulphuric acid is 60.85° Be., 
what is the percentage of H2SO4? (6) What is the Baume of 
78.00% sulphuric acid? 

Ans. The tabular difference in ° Be. is 1° Be. (61 - 60); the 
tabular difference of per cent H2SO4 is 1.76% (79.43% - 
77.67%). The increment in ° Be. of the sample, above the 
given °B6., is 60.85° - 60.00° = 0.85° Be. Assuming that a 
small increase in Baume is accompanied by an increase, in the 
same ratio, of the percentage of H2SO4, the increase of the 
percentage of H2SO4 is 

^ X 1.76 = 1.50%. 

The percentage of H2SO4 in the acid is 

77.67%o + 1.50% = 79.17%. 
(6) Analogously, 

78.00 - 77.67 = 0.33, 

5|| X 1.00 = 0.19. 
1.7o 

60.00 + 0.19 = 60.19° B4. corresponding to 78.00% H2SO4. 

II. Given 

65.75° B^. corresponding to 91.80% H2SO4, 
66.00° B^. corresponding to 93.19% H2SO4. 

If an acid is 66.05° B^., what is its per cent of H2SO4? 

Ans. It may be assumed that the next interval will be in 
approximately the same ratio of increase as in the one previous. 

Tabular difference is 66.00° - 65.75° = 0.25° Be. 
Tabular difference is 93.19%o - 91.80%o = 1.39%, H2SO4. 
66.05° - 66.00° = 0.05° B^. 

^X 1.39 = 0.28% H2SO4. 
93.19 + 0.28 = 93.47% H2SO4 corresponding to 66.05° B^. 



44 



CHEMICAL CALCULATIONS 



43. Given 



°Be. 


Per cent H2SO4. 


Determine 






65 


88.65 


(a) Per cent H2SO4 of 65. 15° 


Be. 


acid 


65i 


89.55 


(5) Per cent H2SO4 of 65.62° 


Be. 


acid 


65i 


90.60 


(c) Per cent H2SO4 of 65.97° 


Be. 


acid 


651 


91.80 


(d) Per cent H2SO4 of 66.06° 


Be. 


acid 


66 


93.19 


(e) °Be. of 89.00% acid 
(/) °Be. of 91.00% acid 
(g) °Be. of 93.35% acid 







Ans. {a) 89.19%; (b) 91.18%; (c) 93.02%; (d) 93.52%; 
(e) 65.09°; (/) 65.58°; (g) 66.03°. 

44. A liquid of 

63° B^. weighs 110.29 lbs. per cubic foot, 
64° B^. weighs 111.65 lbs. per cubic foot. 

What is the weight per cubic foot of a liquid of 63.18° B^.? 

Ans. 110.53 lbs. 

45. Sulphuric acid of 

65.25° Be. corresponds to 96.10% oil of vitriol, 
65.50° Be. corresponds to 97.22% oil of vitriol. 

If a sample has a Baume of 65.37°, what is the percentage of 
oil of vitriol? Ans. 96.64%. 

46. The following is taken from the nitric acid tables of 
Lunge and Rey.^ 



Specific gravity. 


Per cent HNO3. 


1.470 
1.475 


82.90 

84.45 



(a) What is the per cent of nitric acid corresponding to 1.472 
specific gravity? (6) If a sample of nitric acid contains 83.15% 
HNO3, what is its specific gravity? Ans. (a) 83.52%. 

(b) 1.471 sp.gr. 

1 See Chem. Ann., pp. 401-402. 



INTERPOLATION 



45 



47. Ferguson's tables of aqua ammonia furnish the follow- 
ingi; 



Specific gravity. 


Per cent NH3. 


0.8903 

0.8889 


31.85 

32.34 



What is the per cent of ammonia in a sample, if the specific 
gravity is 0.8895? (Notice that as the specific gravity decreases 
the per cent of ammonia increases. These two properties vary 
inversely.) Ans. 32.13%. 

48. Given the following: 



°Be. 


Per cent NH3. 


28.00 

28.25 


33.32 
33.81 



What is the percentage of ammonia of a sample of 28.18° Be.? 

Ans. 33.67%. 
49. Given the following logarithms and the numbers corre- 
sponding : 



Number. 


Logarithm. 


1021 
1022 


3.00903 
3.00945 



(a) What is the logarithm of 1021.6? (b) What is the number 
corresponding to the logarithm 3.00922? Ans. (a) 3.00928; 

(6) 1021.5. 

50. The boiling point of pure water, accurately determined 
in Deer Lodge, Montana, Oct. 25, 1910, was found to be 95.56° C. 
Assuming the thermometer to be correct, what was the baro- 
metric pressure, given the following data? 

1 See Chem. Ann., pp. 408-409. 



46 



CHEMICAL CALCULATIONS 



At 95.00° C. the pressure of aqueous vapor is 633.78 mm. Hg.^ 
At 96.00° C. the pressure of aqueous vapor is 657.54 mm. Hg. 

Ans. 647.09 mm. Hg. 

51. In testing the accuracy of a thermometer, the following 
readings were taken: 

Barometric pressure, 652.70 mm. Hg. 
Reading of thermometer in melting ice, 0.00° C. 
Reading of thermometer in vapor of boiling water, 95.62° C. 
Assuming the bore of the thermometer to be uniform and that 
the barometric pressure exerts no appreciable influence on the 
melting point of ice, what is the error of the thermometer at 
these two points? (Use the data given in Problem 50.) 
Ans. Correct at 0.00° C. 

0.18° C. too low at 95.62° C. 

52. The following readings were obtained on a Jolly balance: 



Load. 


Balance stretches to 


0.040 g. 

0.120 g. 


22.0 cm. 
26.2 cm. 



If the balance stands at 23.7 cm., what load is in the pan? 

Ans. 0.072 gr. 

53. Being given the following data on hydrochloric acid: 
Sp. gr. 1.155 corresponds to 30.55% HCl. 
Sp. gr. 1.145 corresponds to 28.61% HCl. 

(a) Extrapolate the percentage of hydrochloric acid corre- 
sponding to 1.160 sp. gr. (6) 1.140 sp. gr. (c) The specific 
gravity of hydrochloric acid containing 31.25% HCl. {d) 
28.14% HCl. Ans. (a) 31.52%; 

(b) 27.64%; 

(c) 1.159 sp. gr; 

(d) 1.143 sp. gr. 

1 See Chem. Ann., pp. 461-466; also pp. 70-71. 



INTERPOLATION 47 

54. Under a pressure of 760 mm. hydrochloric acid (solution) 
gives a constant boiling point distillate of 20.242% HCl, while 
under 750 mm. the composition of the distillate is 20.266% HCl. 
What is the probable composition of a distillate collected under 
765 mm. Ans. 20.230% HCl. 



CHAPTER IV 
HEAT 

There are four units of heat intensity: 

I. Fahrenheit (written ° F.) . — On the thermometer the 
melting point of water is marked 32°; the boihng point 
(b.p.) of water under standard atmospheric pressure (760 
mm. Hg) is marked 212°. The distance between these 
points is divided into 180 (212 — 32) equal spaces, and the 
same divisions are extended above and below these points. 
The zero of this scale is 32° (intervals) below the melting 
point of ice. A change of heat intensity which will cause 
a change of volume of the substance, usually mercury, 
through one interval of the scale is a change of one degree 
Fahrenheit. This scale is used largely in English-speak- 
ing countries.^ 

II. Centigrade or Celcius (written ° C). On the 
thermometer the melting point (m.p.) of ice is marked 0°; 

^ The expansion of the liquid in the capillary tube is made a 
measure of heat intensity. The mercurial and other liquid ther- 
mometers are based upon the assumption that a change in tempera- 
ture of any number of degrees is accompanied by a change in the 
volume of the liquid by a proportionate amount. Through small 
intervals this is true; the coefficient of expansion of the liquid 
is sensibly uniform. Mercury, being a liquid, does not exhibit a 
uniform expansion for a given range of temperature, at all tem- 
peratures. Gases, however, more closely approximate a uniform 
coefficient of expansion, and for this reason the expansion coefficient 
of hydrogen has been adopted as the standard of temperature meas- 
urement. The hydrogen thermometer and the mercury thermometer 
agree closely in the moderate ranges. See Chem. Ann., pp. 539-541. 

48 



HEAT 49 

the boiling point (b.p.) of water under standard conditions 
of pressure (760 mm.) is marked 100°. The interval is 
divided into 100 equal divisions, and the same divisions 
are extended above and below. This scale is used almost 
exclusively in scientific work and is common in most Con- 
tinental countries. 

III. Reaumur (written ° R.). — On the thermometer 
the melting point of ice is marked 0° and the boiling point 
under standard conditions of pressure (760 mm.) is marked 
80°; the interval is divided into 80 divisions. The grad- 
uations are extended above and below these points as with 
the previous scales. This scale is little used except on 
the Scandinavian Peninsula, in Russia and in some in- 
dustries. 

IV. Absolute (written ° A.). — This scale was derived 
from observations of the changes of volume of gases ac- 
companying changes of temperature. The scale interval 
is the same as in the Centigrade system, but the zero is 
273° below the melting point of ice. This scale is chiefly 
used in gas calculations. 

The calorie, also called the gram-calorie, is the amount 
of heat required to raise the temperature of one gram of 
water one degree Centigrade.^ This unit is very small and 
a larger unit, the big Calorie,^ also called the kilogram- 
calorie (written with a capital C), equal to 1000 gram- 
calories is often employed.^ 

The British Thermal Unit (written B.T.U.) is the 
amount of heat required to raise the temperature of one 

1 See Chem. Ann., p. 481. 

2 Ibid., p. 481. 

2 The value of the calorie varies slightly according to the tempera- 
ture at which it is determined. The amount of heat required to raise 
one gram of water, at 0° C, one degree Centigrade is not the same as 
is required to raise the same amount of water at some other tempera- 
ture, e.g., 15° C, one degree Centigrade. 



50 CHEMICAL CALCULATIONS 

pound of water at 39.1° F. through one degree Fahren- 
heit.i 

Specific Heat. — The number of heat units absorbed 
by a gram of a substance when its temperature is raised 
through one degree Centigrade or given up when it falls 
through one degree Centigrade is called the specific heat 
of that substance.2 

When heat passes from one body to another, the amount 
of heat energy (calories or B.T.U.) gained by the one 
body is equal to the loss of heat sustained by the other. 

The Heat of Fusion (H.F.) is the number of heat 
units required to change a gram of the substance from the 
solid to the liquid state without any change of temper- 
ature. In the case of ice it requires 80 calories to change 
one gram of ice at 0° C. to water at the same temperature. 
The heat of fusion^ of ice is therefore 80. 

The Heat of Vaporization (H.V.) is the quantity of 
heat required to change a gram of the substance from the 
liquid to the gaseous state without any change of temper- 
ature. In the case of water it is 536.6 calories per gram. 

1 See Chem. Ann., p. 481. 

2 The specific heat of a ^substance is a ratio of the heat capacity 
of a unit weight of the substance in question and the heat capacity of 
the same weight of water which is taken as unity, and is independent 
of the units of heat measurement employed. The specific heat of 
magnesium is 0.250; the heat capacity of one pound of magnesium is 
0.250 B.T.U. ; of one gram of magnesium is 0.250 calories; of one kilo- 
gram of the same substance is 0.250 Calories or kilogram-calorie. 

For a tabulation of the specific heats of the elements, see Chem. 
Ann., pp. 4-9. The value of the specific heat of a substance varies 
with the temperature at which it is measured. In obtaining an- 
swers to problems, the specific heat will be assumed to be independ- 
ent of the temperature of determination. 

3 This term is synonymous with "Latent Heat of Fusion" used 
by many writers. The term "Latent" is tautologous in view of 
the definition of the Heat of Fusion. 



HEAT 51 

The Heat of Combustion of a substance may be given 
in the British or Centigrade systems.^ In the Centigrade 
system it is the number of large or small calories evolved 
from the combustion of a kilogram or a gram of the sub- 
stance. In the British system it is the number of B.T.U. 
evolved from the combustion of one pound of the sub- 
stance.2 

Conversion of Centigrade into British System Units 
and Vice Versa. — It follows from the definitions of the 
different units of temperature/ that to convert 

° F. to ° C, solving | (° F. - 32), gives ° C. 
° C. to ° F., solving | ° C. + 32, gives ° F. 
° F. to ° R., solving f (° F. - 32), gives ° R. 
° R. to ° F., solving | ° R. + 32, gives ° F. 
° C. to ° R., solving | C, gives ° R. 

° R. to ° C, solving | ° R. gives ° C. 

° C. to ° A., solving ° C. + 273, gives ° A. 
° A. to ° C, solving ° A. - 273, gives ° C. 

To convert Fahrenheit or Reaumur to Absolute, first con- 
vert to Centigrade and then convert to Absolute. 

1 See Chem. Ann., pp. 481, 502-511. 

2 In American and British mechanical engineering practice, heats 
of combustion are generally given in B.T.U. This is not considered 
the best practice by metallurgical engineers, who prefer the kilo- 
gram-calorie by reason of its obvious advantages when used in con- 
junction with the metric system of weights and volumes. 

3 The interval between the melting point of ice and the boiling 
point of water is 180° for the Fahrenheit scale, 100° for the Centi- 
grade scale and 80° for the Reaumur scale. The zero of the Centi- 
grade and Reaumur scales corresponds to 32° in the Fahrenheit. 
Hence 

F.-32 ^ _C^ ^ R . 
180 100 80* 

Solving yields the formulas given above. 



52 CHEMICAL CALCULATIONS 

From definition, knowing 1 lb. = 453.59 g. and 1° F. = | ° C. : 
1 B.T.U. = 453.59 g. of water raised | ° C. = 453.59 X f 

= 252 cal. = 0.252 Cal. 
1 Cal. = 1000 g. of water raised 1° C. = 1000 cal. = 1.000 Cal. 
Then the factor, given Calories to find B.T.U., is 

and the factor, being given B.T.U. to find Calories, is 
0.252 as found above. 

The heats of combustion may be converted as follows: 
Knowing 1 kilogram = 2.2046 lbs., then if the combus- 
tion of a kilogram of substance produces X Calories, the 
fact may be expressed: 

1 kilo. = X Cal. 

or 2.2046 lbs. =XCal.; 

X Cal. 



then for 1 lb. 



2.2046 



and by the factor for the conversion of Calories into 
B.T.U., found above, the combustion of one pound pro- 
duces 

X 3.96832 _ 
2.2046 - 1-8000 X, 

therefore the factor for the conversion of the heat of 
combustion of a substance expressed in Calories into the 
heat of combustion expressed in B.T.U. is 1.3. Con- 
versely, given the heat of combustion of a substance ex- 
pressed in B.T.U., the factor to convert into Calories is 
1 



1.8000 



= 0.55556.1 



1 A distinction between B.T.U. and Cal. when used in expressing 
a definite amount of heat and v/hen expressing a heat of combustion 
must not be overlooked, as these are two different kinds of quan- 
tities. 



HEAT 53 



PROBLEMS 



I. Convert into Centigrade the following temperatures: 

(a) 80° F. (6) 28° R. (c) 250° A. 

Ans. (a) ° C. = f (80 - 32) = f (48) = 26.66° C. 

(b) ° C. = f X 28 = 35° C. 

(c) ° C. = 250 - 273 = - 23° C. 

II. Convert into Fahrenheit the following temperatures: 
• (a) 60.5° C. (b) 40° R. (c) 273° A. 

Ans. 
(a) ° F. = f X 60.5 + 32 = 108.9 + 32 = 140.9° F. 
(6) ° F. = f X 40 + 32 = 90 + 32 = 122° F. 
(c) ° C. = 273 - 273 = 0° C; ° F. = | X + 32 = 32° F. 

III. (a) How many calories must have been imparted to 
500 g. of water if its temperature is raised from 12° C. to 26° C? ^ 
(b) How many British thermal units? (c) How many Calories? 

Ans. 

(a) 26° - 12° = 14° raise in temp. 

500 X 14 = 7000 cal. 

(b) Factor to convert Calories into B.T.U. = 3.96832, 

7.000 X 3.96832 = 27.778 B.T.U. 

(c) 7000 cal. = |n§ = 7.000 Cal. 

IV. 200 g. of lead at 100° C. are plunged into 200 g. of water 
at 0° C, and the resulting temperature is 3.17° C; what is the 
specific heat of lead? ^ 

Ans. 

200 (3.17 - 0) = 634 cal., heat capacity of the lead, 

200 (100 - 3.17) = 19,366 cal., heat capacity of the water; 

1 Although the value of the calorie varies slightly according to 
the temperature at which it is defined, this variation will not be 
considered in the problems in this book. The calorie will be assumed 
to have a constant value independent of the temperature of defi- 
nition. 

2 In calculating the specific heats, it will be assumed that the 
specific heat of the substance under consideration is independent 
of the temperature, within ordinary ranges. 



54 CHEMICAL CALCULATIONS 

then as the specific heat is, by definition, the ratio of the heat 
capacity of the substance compared with the heat capacity of 
water, 

Or the problem may be solved : 
Let X = the specific heat of the lead, 

then 200 X (100 - 3.17) = 200 (3.17 - 0); 

solving, X = 0.0327. 

V. If 10 g. of ice at 0° C. are brought into contact with 200 g. 
of water at 30° C, what is the temperature of the resulting 
mixture? 

Ans. 

Let X = final temperature. 

80 X 10 = cal. used to melt the ice without any rise, in 
temp, of the resulting water. 
10 X F= cal. used in raising the temp, of this water from 
0° C. to X°. 
200 (30 — X) = cal. given up by the water to melt the ice, and 
to raise the resultant water to temp. X°. 
Then as the heat absorbed is equal to the heat given up, 
(80 X 10) + 10 X = 200 (30 - X), 
X = 24.'8°C. 

VI. In the combustion of one gram of a substance, the tem- 
perature of 700 g. of water was raised 11.03° C. (a) How many 
Calories were liberated? (&) How many Calories per kilo- 
gram? (c) How many Calories per pound would the substance 
liberate? {d) How many British thermal units per pound? 

^Z '^^^^^ = 7.721 Cal. 

(6) 7.721 X 1000 = 7721 Cal. per kilo. 

(c) 1 lb. =453.59 g.; so, 7.721 x453.59 = 3502.2 Cal. per lb. 

(d) Factor to convert Cal. to B.T.U. = 3.96832, 
3502.2 X 3.96832 = 13,898 B.T.U. per Ib.^ 

1 It will be noticed that in this problem, starting with the heat 
given by the combustion of one gram, we have determined the heat 
of combustion in the British system. 



HEAT 55 

55. The melting point of yellow phosphorus is 44.2° C, and 
its boiling point is 290° C. What are the corresponding tempera- 
tures in Fahrenheit? Ans. m. p. 111.6° F., 

b. p. 554° F. 

56. Water is at its greatest density at 4° C. What is this 
temperature in Fahrenheit? Ans. 39.2° F. 

57. The average factory temperature is assumed to be 60° F. 
To what does this correspond in Centigrade? Ans. 15.56° C. 

58. Sir James Dewar reached a temperature of —264° C. by 
evaporating liquid hydrogen under diminished pressure. What 
is this temperature in Fahrenheit? Ans. —443.2° F. 

59. Zero of the Absolute scale is —273° C. What is the cor- 
responding temperature in Fahrenheit? Ans. —459.4° F. 

60. Benzaldehyde melts at 7.7° F. and boils at 355.82° F. 
Find the corresponding temperatures in Centigrade. 

Ans. m. p. —13.5° C, 
b. p. 179.9° C. 

61. If the melting point of platinum is 1753° C, what is this 
in Reaumur? In Fahrenheit? Ans. 1402° R., 

3187° F. 

62. Caproic acid melts at —5.2° C. and boils at 205° C. 
What are the corresponding temperatures in Reaumur? 

Ans. m. p. —4.16° R., 
b. p. 164° R. 

63. Carbon monoxide melts at — 168.8° R. and boils at 
— 152° R. What are the corresponding temperatures on the 
Fahrenheit scale? Ans. m. p. -347.8° F., 

b. p. -310° F. 

64. If —271° C. is the lowest temperature so far obtained, 
what is this temperature in the Absolute scale? Ans. 2° A. 

65. Given the following: 

(a) Hy drogen, critical temp. — 240°C.; b.p. at atm. press. ,—251°C. 
(6) Carbondioxide, critical temp. 31° C; b.p. atatm. press. ,—80°C. 
(c) Alcohol, critical temp. 243° C; b. p. at atm. press., 78° C. 



56 CHEMICAL CALCULATIONS 

What are these temperatures in the Absolute scale? 

Ans. (a) Crit. temp. 33° A., b. p. 21° A.; 
(6) Crit. temp. 304° A., b. p. 193° A.; 
(c) Crit. temp. 516° A., b. p. 351° A. 

66. (a) At what point is the reading on the Centigrade scale 
and the Reaumur scale the same? (6) Centigrade and Fahrenheit? 

Ans. (a) 0°; 
(6) -40°. 

67. 100 g. of water at 5° C. are mixed with 100 g. of water 
at 28° C. What is the temperature of the resultant mixture? 

Ans. 16.5° C. 

68. What is the resultant temperature when 100 g. of water 
at 8° C. are mixed with 250 g. of water at 30° C? 

Ans. 23.7° C. 

69. On mixing the following quantities of water, 

1 kHogram at 30° C. 
1.5 kilograms at 25° C. 
3 kilograms at 20° C. 
4.5 kHograms at 10° C. 

what is the temperature of the resultant? Ans. 17.25° C. 

70. What temperature will result from mixing 15 g. of ice at 
0° C. with 300 g. of water at 30° C? Ans. 24.8° C. 

71. Equal weights of water and ice at 0° C. are mixed, with 
the result that the mixture had a temperature of 5° C. What 
was the temperature of the water? Ans. 90° C. 

72. If 10 g. of steam at 100° C. are condensed in 100 g. of 
water at 15° C, what is the resultant temperature? ^ 

Ans. 71.5° C. 

1 Authorities differ in the values assigned to the heats of fusion 
and the heats of vaporization. This may be owing to different 
temperatures being taken in defining the calorie. The value 80.0 
calories will be taken as the heat of fusion of ice in obtaining the 
answers to the problems in this book. Also 536.6 calories will be 
taken as the heat of vaporization of water. 



HEAT 57 

73. 150 g. of copper at 100° C. are mixed with 200 g. of water 
at 12° C. and the resulting temperature is 17.8° C. What is 
the specific heat of copper? Ans. 0.094. 

74. 200 g. of mercury at 94° C. are mixed with 200 g. of 
water at 17° C. and the resulting mixture has a temperature of 
19.49° C. What is the specific heat of mercury? 

Ans. 0.0334. 

75. The specific heat of iron is 0.113. If 100 g. of water at 
18° C. are mixed with 150 g. of iron at 65° C, what is the re- 
sultant temperature? Ans. 24.8° C. 

76. When one gram-molecule of carbon combines with two 
gram-molecules of sulphur to form carbon disulphide, 19,000 
calories are absorbed. How many British thermal units are 
absorbed? Ans. 75.39 B.T.U. 

77. When two gram-molecules of hydrogen unite with one 
gram-molecule of oxygen to form water, 68,300 calories are 
evolved. To how many British thermal units does this cor- 
respond? Ans. 271 B.T.U. 

78. In the reaction 

CaO + 2 HCl = CaCl2 + H2O, 
183.45 British thermal units are evolved. To how many Calo- 
ries does this correspond? Ans. 46.23 Cal. 

79. The reaction 

NH3 + HCl = NH4CI 
liberates 166.27 B.T.U. How many Calories does this corre- 
spond to? Ans. 41.9 Cal. 

80. By the combustion of a certain mass of coal 200 g. of 
water had its temperature increased from 10° C. to 18.5° C. 
How many Calories were absorbed? Ans. 1.7 Cal. 

81. If the combustion of 7 g. of a substance raises the tem- 
perature of 500 g. of water from 9° C. to 17.4° C, how many 
calories are yielded per gram? Ans. 600 cal. 

82. The heat of combustion of Black Mountain coal is 8333 
Calories. What is this in the British system? 

Ans. 15,000 B.T.U. 



58 CHEMICAL CALCULATIONS 

83. The heat of combustion of a sample of coke was found 
to be 8006 Calories. Convert this into British thermal units. 

Ans. 14,411 B.T.U. 

84. The heat of combustion of a sample of pine was found to 
be 9153 British thermal units. What is the value in Calories? 

Ans. 5085 Cal. 

85. Convert into Calories, 18,718 British thermal units, the 
heat of combustion of a sample of petroleum. 

Ans. 10,399 Cal. 

86. The coefficient of expansion of a gas is 273 of its volume 
at 0° C. for each degree Centigrade of change. What is the 
coefficient of expansion for a degree Fahrenheit? Ans. -^Ij. 

87. One gram of a sample of coal raises 700 cc. of water from 
6.72° C. to 18.26° C. What is its heat of combustion in British 
thermal units? Ans. 14,540 B.T.U. 

88. The combustion of two drams of a substance caused the 
temperature of 9 ounces of water to be raised 15.43° F. (a) 
What is the heat of combustion in the British system? (6) In 
Calories? (16 drams = 1 oz. avoirdupois.) Ans. (a) 1111 B.T.U.; 

(6) 617.2 Cal. 



CHAPTER V 
SPECIFIC GRAVITY 

For the principles underlying this subject, the student 
is referred to textbooks on Physics. 

The density of a body is denned as the weight of a unit 

volume, that is 

, .^ , mass 

density 1 = — -, , 

volume 

from which it follows that 

mass = density X volume 



and 



, mass 

volume 



density 

The relative density, or specific gravity, of a substance, is 
the ratio of its mass, referred to the mass of an equal 
volume of some substance taken as standard. The 
standard substance for solids and liquids is pure water 
taken at its maximum density (4° C. or 39.2° F.). Gases 
have such a low specific gravity referred to water that the 
weight of a unit volume of a gas is usually referred to 
the weight of the same unit volume, say either a liter of 

1 The density of a body is the weight of a unit volume. The 
specific gravity, or the synonymous term, relative density, is the 
ratio of the density of the body in question, referred to the density 
of some substance which is taken as unity. In the metric system, 
the standard employed is water at 4° C. Hence, in this system, the 
density is numerically equal to the specific gravity (or the relative 
density). The formulas given are applicable only when the units 
of measurement are in the metric system. 

59 



60 CHEMICAL CALCULATIONS 

hydrogen, 3V oxygen, or of air. The specific gravity of 
gases will be taken up under Gas Calculations. 

Specific Gravity of Body Heavier than and Insoluble 
in Water. — The relative density, or specific gravity, is 
determined according to the following formula, let 
Wb = weight of the body in air,^ 
Ww = weight of an equal volume of water at 4° C, 

^^"^ w, 

rel. dens, (or sp. gr.) = =^ • 

According to the principle of Archimedes,^ let 

Wb = weight of the body in water at 4° C, 
then 

W^ = Wb- Wb', 
in which Wb — Wb is the loss in weight in water of a 
substance heavier than water {i.e., the body is wholly 
immersed) . Hence 

rel. dens. = ^^^ =^^-7 > 

Wb — Wb 

so, to obtain the relative density (or specific gravity), it 

is only necessary to weigh the substance in air and in 

water and solve by the formula. 

To obtain the relative density when the body is weighed 

in water at some other temperature than 4° C, say 

f C, then 

Wb 
rel. dens. = ^^ — ^^7-. X rel. dens, of water at ^° C. 
Wb—Wb 

^ Strictly, this weight should be corrected to weight in vacuo. 
The reduction of the weight in air to the weight in vacuo is treated 
on pages 102-104 of this book. For a fuller discussion, see Chem. 
Ann., pp. 519-530. 

2 "A body wholly or partly immersed in a fluid (liquid or gas) 
is buoyed up with a force which is equal to the weight of the volume 
of the fluid which the body displaces." A Text-Book of Physics, 
by Duff. 



SPECIFIC GRAVITY 61 

The specific Gravity of a Solid Substance Lighter than 
Water is obtained by means of a sinker attached to the 
soHd; the combination being of a relative density greater 
than water so that it will sink. Let 

Ws' = weight of the sinker in water, 
W's+b = weight of the sinker with body attached in water. 

Then rel. dens. = ^^^^^_tf',^; 

In practice such a determination is most easily carried 
out by weighing the light substance in air and then weigh- 
ing the light substance in air suspended from the hook 
with the sinker attached to it by a slender thread (the 
sinker being immersed in water) and taking the weight 
of this combination. Finally, the weight of both the 
substance and sinker in water is taken. Let 
[Wb + Ws'] = weight of body in air and sinker in water, 
W\+b = weight of body and sinker, both in water. 

Then rel. dens. = ,^^ ^ ^y^^ ^,^^; 

The Specific Gravity of Powders is determined by 

using a specific gravity bottle, flask or pyknometer. The 
empty bottle is weighed, then filled with water and 
weighed again. This gives the weight of water held by 
the bottle. The bottle is emptied and dried and a con- 
venient amount of powder introduced when the bottle 
and powder are weighed. This gives the weight of the 
powder. The bottle containing the powder is now filled 
to the mark and weighed again. Let 

Wb = weight of powder taken, 
Ww+f = weight of flask full of water, 
Wb+f+w = weight of powder, flask and water. 

Wb 



Then rel. dens. = 



Wb - {Wb+f+^ - W^+f) 



62 CHEMICAL CALCULATIONS 

Pyknometer Method for Liquids. — The relative den- 
sity of liquids cannot be obtained by the means outlined 
above. An obvious method is to compare the weight of 
a given volume of the liquid in question with the weight 
of the same volume of water; account being always taken 
of the temperatures of the two liquids. Then, letting 

Wi = weight of a given volume of liquid, 
Ww = weight of the same volume of water, 

Wi 
rel. dens. = :^^ • 

vV w 

This method is used when extremely accurate results are 
desired. 

Sinker Method for Liquids. — The principle of Archi- 
medes states that a body is buoyed up by an amount equal 
to the weight of liquid displaced. Knowing the amount 
of this buoyancy in water and in the liquid the specific 
gravity of which is to be found, the specific gravity of the 
Hquid can be calculated. The buoyancies are measured 
by the losses in weight of a solid body in the liquid of 
unknown density and in water respectively. Any solid 
(usually a metal) of a specific gravity such that it will 
sink in both liquids and is not acted upon by either of 
the liquids will suffice. Let 

Wb = weight of the solid body in air, 
Wb = weight of the solid body in water, 
Wb^ = weight of the solid body in the liquid whose 
density is to be determined. 

Wb - Wb" 



Then rel. dens. 



Wb - Wb' 



Constant Weight Hydrometers. — Another method 
which is based upon the principle of Archimedes is that 
a body lighter than a liquid in which it is immersed 



SPECIFIC GRAVITY 63 

sinks until it displaces an amount of liquid equal to its 
own weight. Let 

A = area of cross section of a cylindrical floating body, 
h = depth to which the cylinder sinks in water, 
Ww = weight of water displaced. 

In the metric system, the unit of weight, the gram, being 
the weight of one cubic centimeter of water (at 4° C), 
then if A and h are measured in square and linear centi- 
meters respectively. Ah represents a volume of water dis- 
placed which is numerically equal to the weight in grams 
of water displaced. That is 

Wy, = Ah. 

If the same cylinder is next placed in a liquid of a differ- 
ent density, it will sink to such a depth that it displaces 
an amount of liquid equal to its own weight. Let 

d = the density of this liquid, 

h' = the depth to which the cylinder sinks in the liquid, 
Wi = the weight of liquid displaced. 

The liquid displaced is 

Wi = Ah'd. 

Then as equal weights of the liquids are displaced (this 
is the method of the constant weight hydrometer), 
Ww = Wi, and as the density of water is unity, then 

Ah'd = Ah, and d = ji- 

In words, the depth to which a cylindrical floating body 
will sink is inversely proportional to the density of the 
liquid in which it is immersed. This is the principle of 
the hydrometer, which is extensively used commercially. 
Constant Volume Hydrometers. — Still another 
method is the use of the principle of flotation, differing 



64 CHEMICAL CALCULATIONS 

from the former method in that the volume of hydrom- 
eter immersed is constant (that is, that it is immersed to 
the same depth in the different Hquids), the weight of the 
liquid displaced varying and serving as a measure of the 
specific gravity. The instrument employed is called the 
Nicholson's Hydrometer. Let 

Wh+g = weight of hydrometer plus the weight in grams 
necessary to be added for it to sink to the 
mark in water, 

Wh+g = weight of hydrometer plus the weight neces- 
sary to be added for it to sink to the mark 
in the liquid of unknown density. 

The weight of the volume of water displaced by the hy- 
drometer when immersed to the mark is Wh+w', the weight 
of the same volume of liquid displaced by the hydrometer 
when immersed to the mark is Wh+g. This gives the 
weights of equal volumes, hence ^ 

rel. dens. = ^^rr^- 

Wh+-g 

Hydrometers are instruments of glass, weighted at 
their lower end, the upper end terminating in a thin 
cylindrical tube. This thin cylindrical tube is graduated, 
not into lengths, but is cahbrated so as to read specific 
gravity directly.^ As the density of a liquid changes with 
the temperature, the liquid should always be at the tem- 
perature at which the hydrometer was calibrated or proper 
correction made. The usual temperature is 60° F. or 
15.5° C. 

The Baume Hydrometer is an instrument constructed 
as a specific gravity hydrometer, but divided into degrees 

1 This same principle is involved in the Westphal's or Mohr 
balance. 

2 See Chem. Ann., pp. 524-525. 



SPECIFIC GRAVITY 65 

Baume.^ These divisions are arbitrary and unscientific, 
but owing to conservatism and their estabUshed use in 
factories manufacturing heavy chemicals, they are still 
retained. As in specific gravity determinations, the 
density of water is taken as unity, but the Baume scale 
differs from the specific gravity scale in that, as the den- 
sity of a liquid decreases below that of water the degrees 
Baume increase. For liquids heavier than water an in- 
crease in specific gravity is accompanied by an increase 
in degrees Baume; consequently it is necessary to know 
whether the degrees Baume apply to a liquid heavier or 
lighter than water.^ 

Conversion of Baume Readings into Specific Gravity 
and Vice Versa. — The formulas showing the relations 
between specific gravity and degrees Baume for liquids 
heavier than water are (temp. 60° F.) 

°Be. = 145--^^^- 
sp. gr. 

145 

For liquids lighter than water, 

og^ = 140 - 130 X sp. gr. ^ 140 _ ^^^ 
sp. gr. sp. gr. 

q 140 

^P*^^' 130 + °Be.^* 

1 See Chem. Ann., pp. 525-526. 

2 For a tabulation of equivalent degrees Baum6 and specific 
gravity for liquids heavier than water, Ibid., pp. 379-383. For 
liquids Hghter than water, Ibid., pp. 383-387. 

3 The Baume scale as devised by Baume was graduated in the 
following manner: The depth to which a hydrometer sank when 
immersed in a solution made by dissolving 15 parts of pure salt in 
85 parts of distilled water was marked 15, temperature 12.5° C. 
The point to which it sank in pure water was marked 0. The in- 



66 CHEMICAL CALCULATIONS 

Calculation of Density Determinations to Different 
Conditions. — In purely scientific calculations, water is 
taken as standard at 4° C, while in commercial labora- 
tories the standard is often in the neighborhood of 15° C; 
consequently specific gravities determined by these 
standards will not agree. As the temperature of water in- 
creases from 4° C, it expands. The weight being con- 
stant, with increase of volume, the specific gravity is 
lowered. In the case of water this increase of volume 
with rise of temperature is not uniform, but has been 
determined with great care. Knowing the relative den- 
sity of water at various temperatures, the volume of a 
gram is obtained by taking the reciprocal of the density. 
The expansion of liquids and solids being appreciable, 
conditions should always be given with densities. This 
is very important in the case of liquids and differences 

15° 
are appreciable with solids. D -^ C. is to be interpreted 

to mean that the temperature of the substance at the 
time of the determination of its density was 15° C. and 

17.5° 

that the unit of density is water at 15° C; D ' C. 

means that the temperature of the substance was 17.5° C. 

terval was divided into 15 equal parts and the same graduations 
extended. This is for use with liquids heavier than water. The 
hydrometer for Hquids lighter than water was marked at the 
point to which it sank in a ten per cent solution of salt and the point 
to which it sank in pure water was marked 10. The interval was 
divided into ten equal spaces and the same distances extended. 
These solutions will not give the exact relations between specific 
gravity and degrees Baume cited in the formulas above. The makers 
of instruments produced so many so-called Baume hydrometers, no 
two of which agreed, that the Manufacturing Chemists' Association 
adopted as the Baume scale one that will agree with the formulas 
given above, which is now regarded as standard. 



SPECIFIC GRAVITY 67 

and is referred to water at 4° C.^ To convert densities 
of substances taken under one set of conditions into 
equivalent densities of the same substance under other 
conditions requires a knowledge of the coefficients of ex- 
pansion (cubical) of the substance under consideration 
and water, or what comes to the same thing, the variations 
in density of the substance and water with change of 
temperature. For water this has been carefully deter- 
mined and may be obtained from tables ;2 in the case of 
other hquids it must be experimentally determined for 
each individual substance. 

x° x° 

To convert D -5 C. to D -^ C. requires only a knowl- 

y 2; 

edge of the densities of water at 2/° C. and z° C. Thus, let 

_ ^x° p _ weight of unit volume of the body at a;° C. _ a; 

y° ' weight of unit volume of water at y° C. y 

a;° p _ weight of unit volume of the body at 0^° C. _x 

z° ' weight of unit volume of water at 2;° C. z 

d = - , X = dy, 



d' ■= -, X = d'z, 

z 

dy = d'z, 

d=-d\ d' = '^d. 
y ^ 

^ Specific gravities may be corrected for the values of the mass 
factor in vacuo (see Tables of Lunge and Isler, Chem. Ann., pp. 
394-396) or this correction may be neglected (see Tables of Fergu- 
son and Talbot, Ihid., pp. 388-393). For corrections for weighing 
in air, see pp. 102-104 of this book. For examples of specific gravi- 
ties given under specified conditions, see, for example, Aluminum, 
Chem. Ann., p. 100; Antimony Pentachloride, p. 106, etc. 

2 For changes in the density and volume of water with change 
of temperature, see Chem. Ann., pp. 457-460. 



68 CHEMICAL CALCULATIONS 



2° 



To convert D ^ C. into D -5 C. requires a knowledge of 

X X 

the variation in density of the body under consideration 
with changes of temperature.^ Thus, let 
, _j^y°^ _ weight of unit volume of the body at 2/° C. _ 2/ 

x° ' weight of unit volume of water at x° C. x 
,, _ ^ ^ p _ weight of unit volume of the body a.tz°C. _ z 

x° ' weight of unit volume of water at x° C. x 



'-1' 


x = dy, 


'^'-l' 


X = d'z, 


dy = d% 




d = -d\ 

y 


d' = U 
z 


PROBLEMS 



L A sample of bismuth weighed 14.738 g. in air and 13.235 g. 
in water, {a) What is the density of the bismuth? (6) What 
is the weight of a cube of bismuth, 2 cm. on an edge? (c) How 
many cubic centimeters in a kilogram of bismuth? 

Wh 
Ans. (a) Rel. dens. = z— -—,• 

Wb. — Wh 
14.738 - 13.235 = 1.503 g. loss of weight in water. 

14.738 ^ one; 
-^^ = 9.805 sp.gr. 

(6) Mass = rel. dens. X vol. 

Mass = 9.805 X (2)^ = 78.445 g, 

(c) Vol. = ^''^^' 
sp. gr. 

1 For corrections in the case of sulphuric acid, see Chem. Ann., 
p. 391. 



SPECIFIC GRAVITY 69 

II. A sample of cork weighed 2.140 g. in air. A silver sinker 
(specific gravity, 10.53) of 10.000 g. was employed, the combina- 
tion of sinker and cork, in water, weighing 2.274 g. Find the 
specific gravity of the cork. 

Ans. Sp. gr. = ^^ 



Wb + W/ - W's+b 

The sinker will displace a volume of water equal to its own 
volume. The weight of this water will be equal to the loss of 
weight of the sinker, when weighed in water. 

-^ = 0.9497 cc. = 0.9497 g. 
10. oo 

10.00 - 0.9497 = 9.0503 g., weight in water of sinker. 

Substituting in the formula, 

2.14 2.14 



2.14 + 9.0503 - 2.274 8.9163 



0.2400 sp. gr. 



III. A block of pine weighed 6.431 g. in air. With a sinker 
attached to the block by a fine thread, the sinker being in 
water and the block in air, the combination weighed 18.530 g.; 
the combination of both sinker and block in water weighed 
7.635 g. Find the specific gravity of the block of pine. 



Ans. Sp. gr. = 



Wj, 



[Wb-\-Ws']-{-W's+i> 
6.431 _ 6.431 _ ^ -Q^. 
18.53 - 7.635 ~ 10.895 ^•^^^'^• 

IV. Find the specific gravity of a sample of sand, from the 
following data : Weight of sand taken, 4.655 g.; weight of bottle 
full of water, 80.04 g.; weight of bottle containing sand and 
filled up with water, 82.755 g. 

Wb 



Ans. Sp. gr. = 



Wb - {Wb+f+w - W^+f) 
4.655 4.655 



4.655 - (82.755 - 80.04) 1.94 



= 2.399. 



70 CHEMICAL CALCULATIONS 

V. A platinum ball weighed 42.96 g. in air, 40.96 g. in water, 
39.548 g. in sulphuric acid and 41.264 g. in naphtha, (a) Find 
the specific gravity of the sulphuric acid, (6) of the naphtha 
and (c) of the platinum. ^ 

Arts, (a) Sp. gr. = |^^^. 
Wb — Wb 

42,96 - 39.548 ^ 3,412 _ 
42.96 - 40.96 2.00 ~ 

(b) 42.96 - 41.264 _ 1,696 _ „ ^ .„ 
^ ^ 42.96 - 40.96 2.00 ~ 

(c) —^^ =4M6^2148 

^ ^ 42.96 - 40.96 2.00 

VI. (a) Convert specific gravity 1.7957 into degrees Baume. 
(6) Convert 65.25° Baume (heavier than water) into specific 
gravity, (c) Convert specific gravity 0.7692 into degrees 
Baume. (d) Convert 51° Baum^ (lighter than water) into 
specific gravity. 

Ans. 
Substituting in the proper formulas: 

(a) ° Be. = 145 - j^ = 145 - 80.75 = 64.25°. 

^^^^-^^-^54^-^5 = ™^ 

(c) ° Be = ^^^ ~ ^^^ ^ ^•'^^92 ^ 140 - 100 _ 40 _ „o 
07692 0.7692 0.7692 ~ ^ * 

(rf)Sp.gr.= -1M- = 140^ 

^ ^ 130 + 51 181 • 

VII. A liquid shows a specific gravity, D ^^^ C, of 1 3182 

15.56° 

The same liquid shows a specific gravity, Z) -^^ C, of 1.3142. 

15.56° 

1 For a tabulation of the specific gravities of elements and chemical 
compounds, see Chem. Ann., pp. 100-337. 



SPECIFIC GRAVITY 71 

17 5° 
What is the specific gravity, D -~^ C? Density of water at 

17.0 

15.56° C. = 0.999040; at 17.5° C. = 0.998713.1 

Ans. 

To change D i||| C. (= d) to D ^°C. ( = d'). 

y° = 15.56°, then y = 0.999040, 
2° = 17.5°, then z = 0.998713. 

To change D ^° C. to Z) |U C. 
^ 1||! C. = 1.3182 

difference = 0.0040 

20° - 15.56° = 4.44°, change in temperature, 

2;^ = 0.000901, difference per ° C. 
4.44 

17.5 — 5.56 = 1.94, difference in temperature, 

1.94 X 0.000901 = 0.0017. 

1.3186 - 0.0017 = 1.3169 specific gravity!) i^ C. 

89. A steel sphere of 1.90 cm. diameter weighed 28.25 g. 
Knowing that 1 cc. of water at 4° C. weighs one gram, what is 
the relative density of the steel sphere? Ans. 7.866. 

90. A block of wood, 7.49 X 7.46 X 3.78 cm., weighs 152.7 g. 
What is its specific gravity? Ans. 0.723. 

91. The largest diamond found weighed 3200 carats. If in the 
form of a cube, what is the length of an edge? (One carat = 205 
milligrams; sp, gr. of diamond, 3.530.) Ans. 5.706 cm. 

92. A footnote in "Pepys' Diary" mentions a stone shot 
weighing 770 lbs., which fired by the Turks in 1807, struck 

1 See Cham. Ann., pp. 457-458. 



72 CHEMICAL CALCULATIONS 

H.M.S. "Lion." If spherical and composed of granite (sp. gr., 
2.5), what was its diameter? (One cubic foot of water weighs 
62.4 lbs.) Ans. 25.35 in. 

93. Linseed oil has a specific gravity of 0.930. What will it 
weigh per gallon? (One gallon = 231 cubic inches; one cubic foot 
of water = 62.37 lbs.) Ans. 7.758 lbs. 

94. A drum has a capacity of 4 cubic feet. How many pounds 
of ammonia of 0.8917 specific gravity will it hold? (Take the 
weight of one cubic foot of water as 62.37 lbs.) 

Ans. 222.5 lbs. 

95. What is the weight of 15 cubic feet of oil of vitriol, the 
specific gravity being 1.8354? Ans. 1717 lbs. 

96. What is the volume of 100 lbs. of hydrochloric acid of 
1.2003 specific gravity? Ans. 1.335 cu. ft. 

97. A casting of iron weighs 1000 kilograms. Taking the 
specific gravity of iron as 7.23, what is its volume? 

Ans. 138 L. 

98. A platinum wire 7.25 cm. long weighs 1.0762 g. The 
specific gravity of platinum is 21.48. Find the diameter of the 
wire. Ans. 0.938 mm. 

99. 0.0203 g. of gold (specific gravity = 19.32) was plated 
on a cubical brass weight 1.5 cm. on an edge. What is the 
thickness of the gold? Ans. 0.00078 mm. 

100. What is the radius of a steel sphere (specific gravity, 
7.81) equal in weight to a brass sphere (specific gravity, 8.40) 
1.5 cm. radius? Ans. 1.54 cm. 

101. Faraday estimated that the ductility of gold was so 
high that the gold in four English sovereigns could be drawn 
into a wire long enough to surround the earth. The weight of 
a sovereign is 7.988 g., and it contains 91.66% gold. If a 
quadrant of the earth is 10,000,857 meters, what is the thick- 
ness of the wire? (Specific gravity of gold, 19.3.) 

Ans. 0.0002198 mm. 

102. A casting of iron is suspected of having internal cavities. 
In air it weighs 170^.42 g.; in water 145.60 g. The specific 



SPECIFIC GRAVITY 73 

gravity of cast iron is 7.23. Has the casting any cavities, and if 
so, what is their volume? Ans. 1.25 cc. 

103. In obtaining the specific gravity of a sample of heavy 
spar (BaS04), the following weights were obtained: Weight in 
air, 5.127 g.; weight in water, 3.969 g. What is the relative 
density of the sample? Ans. 4.427. 

104. In obtaining the specific gravity of a brass weight, the 
following readings were obtained : Weight in air, 116.62 g. ; weight 
in water, 102.81 g.; temperature of the water, 20° C. (One gram 
of water expands to 1.001773 cc. at 20° C.) What is the specific 
gravity of the weight? Ans. 8.430. 

105. Find the weight of a cubic foot of water at 60° F., 
knowing that the relative density of water at 60° F. is 0.999050. 

Ans. 62.363 lbs. 

106. Calculate the relative density of a block, given: 

Weight of block alone in air 152 . 7 g. 

Weight of block alone in air, and sinker in water 218.5 g. 
Weight of block and sinker, in water 9 . 5 g. 

Ans. 0.7308. 

107. Find the relative density of gutta-percha from the 
following data: Weight of gutta-percha in air, 4.152 g.; weight 
of sinker in air, 10.450 g.; weight of sinker in water, 7.546 g. ; 
weight of gutta-percha and sinker, in water, 7.405 g. 

Ans. 0.967. 

108. A sample of willow weighed in air 3.820 g. A sinker 
of lead (specific gravity, 11.4) of a volume of 1.632 cc. was 
employed, the combination weighing in water 14.26 g. What 
is the specific gravity of the willow? Ans. 0.5843. 

109. Determine the relative density of a block of wood from 
the following data: Weight of block in air, 3.750 g.; weight of 
lead sinker, 10.000 g.; weight of lead sinker and block under 
water, 8.315 g. (Specific gravity of lead, 11.34.) 

Ans. 0.824. 

110. A platinum ball weighs 19.278 g. in air, 18.382 g. in 
water and 17.643 g. in sulphuric acid, (a) What is the 



74 CHEMICAL CALCULATIONS 

specific gravity of the platinum ball, and (6) of the sulphuric 
acid? Ans. (a) 21.52. 

(6) 1.8248. 

111. A specific gravity flask holds 83.327 g. of alcohol of 
specific gravity 0.8164; 155.79 g. of sulphuric acid, and 120.44 g, 
of potassium hydrate solution, (a) Determine the specific 
gravity of the sulphuric acid, and (6) of the potassium hydrate 
solution. Ans. (a) 1.526. 

(6) 1.180. 

112. A piece of glass weighed 5.236 g. in an- and its specific 
gravity was 3.256. It weighed 3.702 g. in a solution of am- 
monia. Find the specific gravity of the ammonia. 

Ans. 0.9539. 

113. A cylinder sank 54.40 cm. when immersed in gasoline, 
and 39.85 cm., in water. What is the relative density of the 
gasoline? Ans. 07325. 

114. A cylinder was immersed in water at 4° C, and was 
marked 1.000 at the depth to which it sank. It was then 
immersed in a liquid of 1.2083 specific gravity, and the depth 
to which it sank was marked 1.250. The distance between 
these marks was divided into 25 equal spaces. When the 
cylinder was placed in a thnd liquid, it sank to the 1.150 mark; 
what is the specific gravity of this liquid? Ans. 1.125. 

115. One side of a U-tube is filled with glycerine, the other 
with mercury (relative density, 13.6). If 17.4 cm. of mercury 
balance 187.8 cm. of glycerine, what is the specific gravity of the 
glycerine? Ans. 1.26. 

116. A cylinder, when immersed to a certain depth in water, 
weighed 37.93 g. When immersed to the same depth in gasoline 
it weighed 27.55 g. What is the relative density of the gasoline? 

Ans. 0.7263. 

117. Find the specific gravity of the liquid from the following: 

Weight of specific gravity bottle 40 . 327 g. 

Weight of specific gravity bottle and water. . 143.252 g. 
Weight of specific gravity bottle and hquid. . 108.779 g. 

Ans. 0.665. 



SPECIFIC GRAVITY 75 

118. Bunsen gives the following data; from it calculate the 
relative density of calcium. 

Weight of empty bottle 13 . 64 g. 

Weight of bottle filled with naphtha 20 . 275 g. 

Weight of bottle partly filled with naphtha. . . . 16. 650 g. 
Weight of bottle partly filled with naphtha and 

calcium 19 . 150 g. 

Weight of bottle full of naphtha and calcium. . 21 . 576 g. 

Relative density of the naphtha. 0.758 g. 

Ans. 1.58. 

119. A sample of bronze is made up of 31.50% zinc, 3.00% 
tin and 65.50% copper. What is its specific gravity, supposing 
no change in volume occurred in alloying? (Specific gravities: 
zinc, 7.142; copper, 8.93; tin, 7.29.) Ans. 8.226. 

120. A piece of brass weighed 9.0410 g. in water at 4° C. 
and 10.2621 g. in air. The specific gravity of copper is 8.930 
and of zinc is 7.142. What is the per cent of copper and zinc, 
supposing these two metals, only, are present, and no change 
of volume took place in alloying? Ans. 75.00% Cu, 

25.00% Zn. 

121. An alloy of gold (sp. gr., 19.32) and silver (sp. gr., 10.53) 
has a specific gravity of 13.6312. Assuming that no change of 
volume occurs in alloying, what is its percentage composition 
(a) by volume? (b) By weight? 

Ans. (a) 35.28% Au, 

64.72% Ag. 

(b) 50.00%oAu, 

50.00% Ag. 

122. If the volume of the moon is iVth and its mass sist 
that of the earth, (a) what is the density of the moon compared 
to the earth? (6) If the relative density of the earth is 5.53, 
what is the relative density of the moon? • Ans. (a) 0.605. 

(6) 3.34. 

123. An amalgam consisting of 60.34% mercury (specific 
gravity, 13.59) and of 39.66% gold (specific gravity, 19.3) shows 



76 CHEMICAL CALCULATIONS 

a specific gravity of 15.47. What is the contraction in volume 
that has taken place in the formation of a kilogram of the 
amalgam? Ans. 0.31 cc. 

124. Lupton states that an alloy of 50.00% by weight of 
platinum (specific gravity, 21.5) and 50.00% by weight of cop- 
per (specific gravity, 9.00) has the same color and density as 
gold (specific gravity, 19.5). What is the contraction in mak- 
ing 50 cc. of the alloy? Ans. 26.84 cc. 

125. Find the degrees Baume corresponding to the following 
specific gravities: (a) 1.8354; (6) 1.5263; (c) 1.2205; (d) 0.8963; 
(e) 0.9315. Ans. (a) 66° Be. 

(6) 50° Be. 

(c) 2Q.2°B6. 

(d) 26.2° Be. 

(e) 20.3° B6. 

126. Fiad the specific gravities corresponding to the follow- 
ing: Liquids heavier than water; (a) 65.75° Be.; (6) 30.6° Be. 
Liquids lighter than water; (c) 20.85° Be.; (^) 26.92° B^. 

Ans. (a) 1.8297 sp. gr. 

(6) 1.2675 sp. gr. 

(c) 0.9281 sp. gr. 

(d) 0.8922 sp. gr. 

127. The allowance for temperature, of 13% to 26% nitric 
acid, is 0.00029 specific gravity, for each degree Fahrenheit, 
(a) Given a sample of acid of specific gravity 1.1154 at 60° F., 
what is its specific gravity at 45° F.? (6) At 78° F. (c) What 
is the weight of 3.4 cubic feet of this acid at 80° F. {d) What 
weight of this acid will occupy 10 cubic feet, at 42° F.? (e) 
What is the volume in cubic feet of 100 lbs. of this acid 
at 60° F.? (/) At 73° F.? (One cubic foot of water at 60° F. 
weighs 62.37 pounds.) Ans. (a) 1.1197 sp. gr. 

(6) 1.1102 sp.gr. 

(c) 235.3 lbs. 

(d) 698.9 lbs. 

(e) 1.437 cu. ft. 
(/) 1.442 cu. ft. 



SPECIFIC GRAVITY 77 

128. An acid of a certain concentration was found to have a 
specific gravity of 1.5281 at 56° F., and a specific gravity of 
1.5209 at 72° F. {a) What is the average expansion per de- 
gree F. in cubic centimeters? (6) What is the change per 
degree F. of the specific gravity? (c) Change of the strength 
Baume per degree F.? (d) W^hat is the specific gravity of this 
acid at 60° F.? (e) What is the Baume strength of this acid 
at 60° F. (/) Assuming the changes of specific gravity and of 
the Baume strength to vary uniformly with the temperature, 
what is the specific gravity of the acid at 50° F.? (g) What is 
the strength Baume at 80° F.? Ans. (a) 0.000296. 

(6) 0.00045. 

(c) 0.02812. 

(d) 1.5263. 

(e) 50.00° Be. 
(/) 1.5308. 
(g) 49.44°. 

129. 60° F. is the temperature at which degrees Baum^ are 
tabulated. An acid of a certain concentration changes 0.0235° 
Be. for each degree change of temperature (Fahrenheit), (a) 
If the Baume at 42° F. of a sample of this acid is 66.46° Be., 
what is the Baume at the temperature of tabulation? (6) What 
would be the Baume of this acid at 73° F.? (c) If at 60° F. 
the per cent of acid corresponding to 66° B6. is 93.19%, and 
65.75° Be. corresponds to 91.80% acid, what is the per cent 
acid in this sample? Ans. (a) 66.04° B^. 

(6) 65.73° B4. 
(c) 93.41%. 

130. A sample of sulphuric acid is 65.25° Be. at 60° F. How 
many pounds per cubic foot does it weigh? Ans. 113.40 lbs. 

131. What must be the diameter of a drum, to hold 400 lbs. 
of 26° Be. ammonia (ammonia water), length of drum to be 2.5 
feet? (Ammonia is lighter than water.) Ans. 1.91 ft. 

132. Accurate volumetric analysis requires that correction 
be made for changes of volume of standard solutions with 
change of temperature. A solution was standardized at 72° F. 



78 CHEMICAL CALCULATIONS 

This solution showed a specific gravity of 1.0277 at 84° F., and 
1.0378 at 40° F. (a) What is the expansion per unit volume 
per degree Fahrenheit? (6) If a determination was made with 
this solution at 55° F., using 98.00 cc, what correction must be 
made to find what the volume would be at 72° F., the tempera- 
ture at which it was standardized? (c) What is the volume 
corrected to 72° F.? Ans. (a) 0.000225. 

(6) + 0.37 cc. 

(c) 98.37 cc. 

15° 

133. The specific gravity, D — C, of a 70% mixture of alco- 
hol and water (by weight) is 0.87187. The specific gravity, 

20° 
D -75" C, is 0.86766 for the same mixture, (a) Calculate 

17 5° 15° 

D — 7^- C. and (6) D —^ C. for this liquid. Density of water 
4 15 

at 15° C. = 0.999126. Ans. (a) 0.86976. 

(6) 0.87264. 

134. The specific gravity, D ^^ C, of 50.87% H2SO4 is 

lo.oo 

1.4078. (a) What is the specific gravity, D ^^^ C? (b) 

15° 

Z) — C? Sulphuric acid of this concentration decreases 

0.000738 D ^^^ C. for each ° C. Density of water at 15.56° C. 
15. 5d 

= 0.999040. Ans. (a) 1.4065. 

(6) 1.4069. 

135. Nitric acid containing 50.32% HNO3 has a specific grav- 

15 56° 
ity, D ^_' ^- C, of 1.3182. The same acid shows a specific 
15.00 

2G 

15.56 

17.5 

17.5 

17.5° C. = 0.998713. Ans. 1.3165. 



gravity, D 7^-^^ C, of 1.3134. What is the specific gravity, 
D i;^ C? Density of water at 15.56° C. = 0.999040; at 



SPECIFIC GRAVITY 79 

136. The Baume, D ^§^ C, of 77.67% H2SO4 is 60.00°. 

15.00 

20° 
This same acid shows a Baume, D -— — rC, of 59.79°. What 

15.00 

is the Baume, D ^ C? Ans. 59.97° Be. 

137. Calculate the difference in specific gravity in the tables 
of Ferguson and Talbot and those of Lunge and Isler on an 
acid of 50.00% H2SO4, being given the following: 

Ferguson and Talbot 

49.47% H2SO4 = 1.3942, D ]^ C. 

15. 5o 

50.87% H2SO4 = 1.4078, D ]^C. 

15.5o 

Lunge and Isler 
50.11% H2SO4 = 1.4000, D ^C. 

50.63% H2SO4 = 1.4050, D ^C. 



Density of water at 15.56° C. = 0.999040. 
5O4 = 0.000738 D i 
Ans. L. and I., 0.0023 sp. gr. lower. 



Correction for 50% H2SO4 = 0.000738 D i|4^ C. for each ° C. 

15. 5o 



CHAPTER VI 

GAS CALCULATIONS 

Boyle's Law. — The temperature remaining constant, 
the volume of a gas varies inversely as the pressure to 
which it is subjected. Let V be the volume of a gas under 
a pressure P, and let V be some other volume of the same 
quantity of the gas and P' its corresponding pressure. 
The analytical expression of this law is 

V'~ P' 
or 

PV = P'V'} 

Charles' Law. — The pressure remaining constant, the 
volume of a gas varies directly as the absolute temper- 
ature to which it is subjected. Let V be the volume of a 
gas at a temperature T, and let V be some other volume 
of the same quantity of the gas and T" its corresponding 
temperature. Then the analytical expression of this law 
is 

Since 0° C. corresponds to 273° A., the law of Charles 
may be stated : The pressure remaining constant, a true 

1 P'y = fc, a constant, therefore on plotting the changes of a 
given volume of a gas under varying pressure or temperature an 
hyperbola results. 

2 Note that T and T' are in the absolute temperature scale. 

80 



GAS CALCULATIONS 81 

gas expands or contracts 2H of its volume at 0° C, for 
each degree Centigrade rise or fall in temperature.^ 

Furthermore, the volume remaining constant, the 
pressure on a gas varies directly as the absolute temper- 
ature. Let P be the pressure of a gas at temperature T 
and let P' be some other pressure on the same quantity 
of the gas and T' its corresponding temperature. Then 
the analytical expression of this fact is 

p' r* 

The gas thermometer is based upon this law. The 
standard degree of temperature is a temperature interval 
such as will cause the pressure on a confined gas to change 
1 Jo of that change in pressure which is shown by a true 
gas between the temperature of melting ice and the 
temperature of water boiling under standard pressure. 
Thus, the pressure exerted by a gas is used as a means of 
measuring temperature, and is employed in the hydrogen 
thermometer, in which the volume is kept constant and 
differences of pressure caused by different temperatures 
are measured. This unit has been chosen for the reason 
that the expansion coefl^icient of hydrogen is very uni- 
form over wide ranges of temperature, a property of all 
gases in a condition far removed from their liquefaction 
point. Mercury, being a liquid, does not expand with 
this regularity with increase of temperature, although at 

^ 273 may be expressed as a decimal, when it becomes 0.003663. 
Every gas has its own coefficients of expansion, one under constant 
pressm-e, which is the coefficient usually required for gas calculations, 
the other under constant volume, a coefficient of a different numerical 
value. For air the coefficient has been found to be 0.0036706 (under 
constant pressure) and the decimal value 0.00367 is usually employed 
for every gas where extreme accuracy is not required. See Chem. 
Ann., p. 73. 



82 CHEMICAL CALCULATIONS 

ordinary temperatures the difference between a temper- 
ature reading with a hydrogen thermometer and with a 
mercury thermometer is sUght.^ 

Laws of Boyle and Charles Combined. — The laws of 
Boyle and Charles are readily combined in one expression. 
Considering a given weight of gas, Boyle's law is 

Poc— , hence P = K^i 

in which X is a constant. The law of Charles is 

PccT, hence P = ICT, 

in which K' is a constant. Combining these expressions:^ 

n T PV 

P = r y, or -yT = ^ 

in which r is a constant for the same quantity of gas. 
From this it follows that the same mass of gas under the 
conditions P\ V and T' gives 
P'V 

rpf ' f 

hence 

PV ^ P'V 

T " r ' 

^ Gases only conform to these generalizations approximately. 
In general, the fm-ther a gas is removed from its liquefaction tem- 
peratm-e, the more closely does it obey the gas laws. An "ideal" 
gas is a hypothetical gas which is supposed to exactly obey the gas 
laws. A distinction between gases and vapors is sometimes made, 
assigning to the former a condition of temperature above its critical 
temperature, and to the latter a condition of temperature below its 
critical temperature. 

2 If the volume varies inversely as the pressure and directly as 
the absolute temperature, then the product of the volume into the 
pressure is equal to the absolute temperature into a constant. 
When a gram molecular volume (G.M.V.) is under consideration, 
this is usually expressed, PV = RT. 



GAS CALCULATIONS 83 

Knowing five of these quantities, the sixth may be ob- 
tained by solving the equation. It is more to be recom- 
mended, however, that in solving gas equations the logic 
of the case should serve as a guide than that a formula 
should be used unthinkingly. For example, a volume of 
a gas equal to F at a temperature T is to be changed to 
temperature T' , pressure remaining constant. Knowing 
that a gas expands with a rise and contracts with a fall 
of temperature, a ratio may be made employing T and 
T' which is a proper or an improper fraction according 
as the gas contracts or expands. This ratio is multipHed 
by V to obtain V. The same may be said for changes of 
volume of gases with change of pressure, temperature 
remaining constant. Let V be the volume of a gas at 
pressure P, which is to be changed to pressure P' . The 
new volume V will be found by using P and P' as 
terms of a ratio, recollecting that if the new condition 
of the gas is to be a smaller volume than V, the ratio 
must be in the form of a proper fraction; if larger, an 
improper fraction. This fraction is multiplied by V to 
obtain V. 

Standard Conditions of Pressure and Temperature. — 
As changes of temperature and pressure exert so consider- 
able an influence upon the volume of a gas and conse- 
quently upon the weight of a unit volume, standard 
conditions have been adopted, which are: 0° C. as stand- 
ard temperature, and a pressure equal to that exerted 
by a column of mercury 76 cm. high when at a temper- 
ature of 0° C.^ The measurement of temperature has 
already been discussed ; it remains to take up the measure- 
ment of pressure. Pressure is usually expressed as so 
many units of weight per square unit of area. Atmos- 

^ See Chem. Ann., p. 3. 



84 CHEMICAL CALCULATIONS 

pheric pressure is measured by the height of a column of 
mercury which will balance this downward pressure of 
the atmosphere. Then if the height of the mercury 
column is taken to be standard at 76 cm. at a temper- 
ature of 0° C. and the specific gravity of mercury is 
13.596 at this temperature, the pressure per square cen- 
timeter is 13.596 X 76 = 1033.3 g. As atmospheric pres- 
sure is always measured by a mercury barometer or a 
barometer standardized against a mercury barometer, it 
is only necessary to indicate the height of the column of 
mercury. 

Correction of Barometer for Temperature. — Mercury 
expands and contracts with rise and fall of temperature; 
consequently its specific gravity increases and decreases. 
So, merely measuring the height of the column of mercury 
is not an exact measurement of the pressure. As 0° C* 
is the standard of temperature of a gas, so also is this 
same temperature taken at which the height of the 
mercury column is standard. For example, at 0° C. the 
specific gravity of mercury is 13.596, then the pressure 
of 76 cm. of mercury at this temperature is 76 X 13.596 = 
1033.296 g. per square centimeter. At 15° C. the specific 
gravity of mercury is 13.560, then the pressure of 76 cm. 
of mercury at this temperature is 76 X 13.560 = 1030.56 g. 
per square centimeter. The height of the column of 
mercury is usually measured by graduations on the glass, 
or the glass tube is mounted in a brass jacket which 
carries the graduations. If the expansion of the mercury 
and of the substance carrying the graduations were the 
same, no correction for temperature in reading the barom- 
eter would be necessary; because, as the mercury ex- 
panded, though its specific gravity would be lowered, the 
material carrying the graduations would expand by an 
equal amount and these expansions would neutralize each 



GAS CALCULATIONS 85 

other, as the graduations would register a greater length 
than the true length.^ 

As an expansion of the mercury is accompanied by an 
expansion of the material carrying the graduations, it is 
only necessary to determine the difference between the 
coefficient of expansion of the mercury and its containing 
tube. This is the apparent expansion of the mercury, 
and for measuring this height on a glass tube 0.00017 of 
the column at 0° C. for each degree departure from this 
temperature is the correction. For a brass scale it is 
0.00016. Hence 1 zL at is the length of the apparent 
column as compared with the column at 0° C. In this 
expression a is the apparent coefficient of expansion and 
t is the number of degrees from 0° C. Then if P is the 
observed height of the barometer, the corrected reading 

. P 2 

'^ 1 ± at 

Moist Gases. — Volumes of gases are often measured 
over liquids which may or may not exert an appreciable 
vapor pressure. The vapor pressure of a saturated vapor 
depends upon the temperature only, and is independent 
of the pressure or the presence or absence of an inert gas. 
If a sufficient amount of a volatile liquid is introduced 
into a Torricelhan vacuum above the mercury in a barom- 
eter or into a barometer tube containing a gas, the height 
of the mercury column will be depressed by an amount 
which is independent of all conditions except the temper- 
ature. If, then, the volume of a confined gas is measured 

1 For the very accurate reading of some forms of the mercm-y 
barometer, capillarity must also be taken into account. This cor- 
rection will not be considered here as it varies with the form, of the 
instrument and the diameter of the tube. This correction is usu- 
ally supplied by the manufacturer with each instrument. 

2 For a fuller discussion, see Chem. Ann., pp. 542-543. 



86 CHEMICAL CALCULATIONS 

over a volatile liquid such as water, the volume will appear 
greater than the volurae of the same amount of dry gas 
by an amount corresponding to the vapor pressure of the 
water (if this be the liquid employed)^ at that temper- 
ature. If this vapor pressure were a constant quantity 
or increased regularly with rise in temperature, it would 
be a simple matter to correct for it; but such not being 
the case, the vapor pressures corresponding to various 
temperatures are determined experimentally and tabu- 
lated. ^ Let 

Pg+w = the pressure of the moist gas, 
Pw = the pressure of the water vapor, 

Vg+w = the volume of the moist gas, 
Vg = the volume of the dry gas. 
Then Pg^^ — Pw is the pressure of the dry gas, the 
volume of which is 

V — ^S + lO ~ Pw -XT 

y g — n y g + w' 

i^g + w 

When measuring a gas over mercury, whether moist 
or not, a common procedure is to bring the mercury to 
the same level inside and outside the tube, the pressure 
of the gas being that indicated by the barometer. It 
may not be convenient to do this; then, to measure the 
pressure of a gas confined in this way, the height of the 
mercury in the tube must be subtracted from the baro- 
metric reading.^ 

1 Every liquid has a vapor pressure which is peculiar to that 
liquid. To determine the variation of vapor pressure of a liquid 
with change of temperature is a matter of experiment. 

2 For the values for water, see Chem. Ann., pp. 461-466. 

2 If the gas is measured over some liquid other than mercury and 
the level inside and outside the tube is not the same, the height of 
the liquid must be reduced to the equivalent height of a column of 
mercury. This necessitates a knowledge of the specific gravities 
of the mercury and the liquid. 



GAS CALCULATIONS 87 

Let 

Pb = the pressure indicated by the barometer, 
Ph = the height of mercury in the tube. 

Then the pressure of the moist gas is Pb — Ph and the 
pressure of the dry gas is (Pb — Ph) — Piu, or Pb — 
(Ph + Pw), and the volume of the dry gas is 



^ Pb — (Ph -\- Pw) y 1 

g ~ p ^ g + w' 



Gay-Lussac's Law of Combination by Volume. — In 

a chemical combination of gases, producing a gas, there is 
always a simple ratio between the volumes of the factors 
(the gases entering into the reaction) and the volumes 
of the products (the gas or gases resulting). Chemical 
equations represent not only combination by weight, but 
also combination by volume. From the equation 

1 Tables are constructed for reducing the volume of a gas, moist 
or dry, to the volume which the gas would occupy dry at standard 
conditions (0° C. and 760 mm.). For example, 300 cc. of a gas is 
measured moist at 18° C. and 765 mm. What is the volume, dry, 
at 0° C. and 670 mm.? Using the decimal coefficient, the expres- 

don to solve is V, = i + (13 x 0.00367) ^ ^"^^i)^ ^ ^°°- ™" 

logarithm of the factor ^„„.., — , ^ ^^o/^>7 ^a ^^^d the values of Pw for 
760(1 + 0.00367 t) 

different temperatures are given in the table. To solve by use of 

the table: 

■°g 760(1+ 0.00367 /°^^^°^- = ^'^^l" 

Ten. aq. vap. for 18° C. 

= 15.38 mm.; 765 - 15.38 = 749.62; log = 2.87484 

log 300 = 2.47712 
sum of logs = 2.44337 
antilog = 277.57 
This considerably simplifies the calculation. See Chem. Ann., 
pp. 70-71. 



88 CHEMICAL CALCULATIONS 



2H20 = 


2H2 


+ 


O2 


36.032 


4.032 




32 


2 vols. 


2 vols. 




1 vol. 



(a) 
(6) 

the subscripts (a) give the decomposition by weight, 
while subscripts (h) indicate the volumes entering into the 
reaction. It is to be noted that in writing equations from 
which calculations involving volumes are to be made, care 
must be taken to represent the molecular condition of the 
gases. When this is done the coefficients of the symbols 
indicate the volumes of factors and products. 

Gas Analysis. — Gay-Lussac's law finds constant ap- 
plication in gas analysis. Given a mixture of hydrogen, 
carbon monoxide and methane; the mixture may be 
analyzed by exploding the whole with an excess of oxygen 
and noting the contraction. The resulting carbon dioxide 
is next absorbed in potassium hydroxide and the contrac- 
tion again noted. The reactions taking place are: 



(a) 


2H2 


+ 


02 


= 2H2O 




2 vols. 




1vol. 


vol. 


(b) 


2 00 


+ 


02 


= 2 CO2 




2 vols. 




1vol. 


2 vols. 


(c) 


CH4 


+ 


202 


= CO2 + 2H2O 




1vol. 




2 vols. 


1 vol. vol. 



In regard to the contraction caused by oxidation, the 
following is to be noted : 

Reaction (a) shows that two volumes of water vapor 
and one volume of oxygen, in all three volumes, disappear 
in the oxidation of two volumes of hydrogen (the water 
vapor being condensed to a liquid, the volume of which 
may be neglected, being approximately only jyVo of the 
volume it occupies as steam). Then, if x represents the 
volume of hydrogen originally present, | x is the amount 
of contraction due to the oxidation of Ijydrogen. 



' GAS CALCULATIONS 89 

Reaction (6) shows that where originally there were 
three volumes of gas present before the explosion, two 
volumes of carbon monoxide and one volume of oxygen, 
after oxidation there remain only two volumes of carbon 
dioxide, a net loss of one volume. If y represents the 
volume of carbon monoxide originally present, \ y repre- 
sents the contraction due to the oxidation of carbon 
monoxide. 

Reaction (c) shows that where originally there were 
three volumes, 07ie volume of methane and two volumes 
of oxygen, after explosion there remains but one volume, 
the loss in this case being two volumes. As this contrac- 
tion was caused by one volume of methane, then, if z 
represents the volume of methane originally present, the 
contraction due to methane is 2 z. 

After exploding the mixture, the resulting carbon 
dioxide is absorbed by potassium hydroxide, the reaction 
being 

CO2 + 2 KOH - K2CO3 + H2O. 

Reaction (a) produces no carbon dioxide, hence no 
part of the carbon dioxide absorbed can be ascribed to 
hydrogen. 

Reaction (5) shows that carbon dioxide to the extent 
of two volumes will be absorbed by the potassium hydrox- 
ide, and as these two volumes of carbon dioxide resulted 
from two volumes of carbon monoxide, it is evident that, 
if y be the volume of carbon monoxide originally present, 
the same volume will be absorbed. 

Reaction (c) shows that one volume of carbon dioxide 
is absorbed which was produced from one volume of 
methane. If z represents the volume of methane origi- 
nally present, the contraction, due to the absorption of 
carbon dioxide produced from z volumes of methane, will 
be z. 



90 CHEMICAL CALCULATIONS 

To put these facts in algebraic form, let 

A = volume of mixture taken, 

B = volume of contraction after explosion, 

C = volume of carbon dioxide absorbed, 

X = volume of hydrogen, 

y = volume of carbon monoxide, 

z = volume of methane. 

(a) A = x-\-y-{-z. 

(6) B = ^x + iy-\-2z. 

(c) C = y + z. 

(a) A = x + y + z. 

(c) C = y + z. 

(d) A — C = X. (by subtraction.) 

Substituting the value of (d) in (a), clearing (6) of frac- 
tions and solving: 

(e) A=A-C + y-^z. 

(J)) 2B = 3A - 3C + i/ + 4g. 

A-2B=-2A-\-2C-Sz. (by subtraction.) 

... -3A+25 + 2C 
(/) z= 3 

Substituting (/) and (d) in (a) and solving for y : 
A=A-C + y + ^^A±lB + 2C^ 

3A = SA-SC + Sy-3A-\-2B-\-2C. 
SA-2B + C 

y= — 3 

Avogadro's Law. — Equal volumes of gases, elemen- 
tary or compound, at the same temperature and pressure, 
contain the same number of molecules. 

This law gives a method of determining molecular 
weights. Let n be the number of molecules of gas A, and 



GAS CALCULATIONS 91 

m the weight of each molecule, then the total weight w 
of this gas is 

w = nm. 

Also let m' be the weight of each molecule of another 
gas B of the same volume and under the same condi- 
tions of temperature and pressure, then the weight w' 
of this gas is 







w' = 


■ nm'] 


the ratio of the 


weight 


w to 


w' is 






w 


nm 






w'~ 


nm' ' 


consequently, 












w 


m 
'm'' 



or the ratio of the molecular weights of two gases is the 
ratio of the weights of equal volumes under the same con- 
ditions. So if 5 is the gas taken as standard, the molecular 
weight m of the gas A is w, or the weight of a unit volume 
of the gas A referred to the weight of the same volume 
in the same condition as the gas B. 

Atomic Weights and Molecular Weights. — Hydro- 
gen was the gas formerly taken as the basis of atomic 
weights; but lately an ideal gas of a weight 3V of oxygen 
gas has come into almost universal acceptance. For 
practical purposes air is often used in determining molec- 
ular weights. From- what has been stated it is evident 
that these gases bear the same relation to the determina- 
tion of molecular weights as water bears to the deter- 
mination of specific gravity. As one cubic centimeter 
of water at 4° C. is taken as the unit weight of a unit 
volume in specific gravity determinations, so in molecular 
weight determinations, which are analogous, the unit 



92 CHEMICAL CALCULATIONS 

of weight is the weight, under standard conditions, of one 
liter of hydrogen, -^^ of the weight of oxygen gas, or, lastly, 
the weight of one liter of air. 

Atomic weights are now based on oxygen as 16, and as a 
molecule of oxygen is composed of two atoms, the molec- 
ular weight of oxygen is 32, so the unit in atomic weights 
is ^V of the weight of oxygen gas.^ 

One liter of oxygen, at sea level, latitude 45°, under 
a pressure of 760 mm. of mercury and at 0° C. weighs 
1.4290 g. Hence one liter of an imaginary gas having a 
unit atomic weight would weigh, per liter, 

1.4290 ^n/i^Af^A 
g2 = 0.044656 g. 

Consequently, if m is the molecular weight of any gas, 
0.044656 m = weight of a liter under standard conditions, 
and, conversely, given the weight ly of a liter of gas under 
standard conditions, the molecular weight m is 

w 
m 



0.044656 



Instead of dividing the weight of a liter of the gas by 
0.044656 it may be multiplied by the reciprocal 22.393. 
In general terms the relation .between the weight w oi a. 
liter of gas and its molecular weight m is 

w = ^X 1.4290 = 0.044656 m g. 

Vapor Density. — Relative density being defined as 
the ratio of the weight of a unit volume of the substance 
under consideration referred to the weight of the same 

1 Most of the common elementary gases have two atoms in the 
molecule. Examples; CI2, H2, N2, etc. See Chem. Ann., p. 3. 



GAS CALCULATIONS 93 

volume of another substance taken as standard/ the for- 
mula for relative density is 

Density to standard substance X 
_ wt. of unit vol. of substance 
wt. of unit vol. of X 
The standard substance for solids and liquids is water at 
4° C. ; for gases 
Density to gV oxygen ^ 

_ wt. of one liter of gas 

wt. of one liter of ideal gas, ^^2 O2 (0.044656) ' 

-^ ., , . wt. of one liter of the gas 

Density to air = — - — ^ ^ . g . , 

wt. of one hter of air (1.2926) 

Density to hydrogen 

_ wt. of one liter of the gas 

~ wt. of one liter of hydrogen (0.089873) ' 

hence it is evident that given the specific gravity (vapor 
density) of a gas to hydrogen, to obtain the molecular 
weight m, divide the weight to of a liter by 2.016, the 
molecular weight of hydrogen: 

0.089873 ^^,,^on ^u r ^ 

0.044580; therefore m 



2.016 ' .^.^.^.^ 0.04450 

Air weighs 1.2926 g. per liter under standard conditions; 
considering it to be a chemical compound it has a molecu- 
lar weight referred to ^V oxygen of 

1.2926 



0.044656 



= 28.946. 



1 When the metric system is employed, the specific gravity and 
the density of a gas is numerically the same if the unit of volume 
is the liter. As already mentioned, the terms relative density and 
specific gravity are synonymous; when deafing with gases the term 
vapor density is frequently employed. 

2 The specific gravity to 3^2 oxygen is the molecular weight. 



94 CHEMICAL CALCULATIONS 

Consequently, given the specific gravity g oi a gas referred 
to air, the molecular weight m is 

m = gX 28.946, 

and, conversely, given the molecular weight, its specific 
gravity is 

_ m 
^ ~ 2SM6' 

The density of a gas (pressure remaining constant) is 
inversely proportional to the absolute temperature, that 
is, 

D^ r. 

also the density of a gas varies directly as the pressure 
(temperature remaining constant), or 

Graham's Law. — The volumes of two gases which 
diffuse in equal times under the same conditions are in- 
versely proportional to the square roots of their densities. 
Effusion, or the passage of gases through a minute orifice 
in a thin plate, obeys the same law as diffusion, which is 
the passage of gases through a porous diaphragm. The 
analytical expression of this law is 

in which Vi is the volume of a gas of density Di which 
diffuses or effuses in the same time as a volume F2 of 
another gas of density D2. 

Bunsen found that the ratios of the times of diffusion 
(or effusion) of two gases which will diffuse (or effuse) the 



GAS CALCULATIONS 



95 



same volume is directly proportional to the square roots 
of their densities; that is, 



T, 



vm' 



or 



in which Di is the density of a gas that requires time Ti 
to diffuse (or effuse) the same volume as the gas of density 
D2 which requires time T2. 

Volume Occupied by a Gram Molecule of a Gas. — 
A gram molecule of a substance will be defined as that 
weight of the substance, in grams, numerically equal to its 
molecular weight. For example, the molecular weight of 
oxygen is 32, hence a gram molecule of oxygen is 32 g. 
Given the following : 



Substance. 


Molecular weight. 


(a) CI2 
(6) HCl 
(c) CO2 


70.92 

36.468 

44.00 



to find the volume occupied by one gram molecule of each 
of these substances. 

(a) 70.92 X 0.044656 = 3.167 g. per liter, 
70.92 



3.167 
or in one expression 
70.92 
70.92 X 0.044656 
(6) Similarly, 
36.468 
36.468X0.044656 
(c) In the same way, 
44.00 
44.00 X 0.044656 



= 22.393 liters per gram molecule; 
= 22.393 liters per gram molecule. 

= 22.393 liters per gram molecule. 

= 22.393 liters per gram molecule. 



yo CHEMICAL CALCULATIONS 

A consideration of Avogadro's law would have led to 
these results, for since all equal volumes of gases under 
the same conditions contain the same number of mole- 
cules, and as this fact serves as a means of determining 
molecular weights, it follows that if weights of gases are 
taken according to their molecular weights, these gases 
must occupy the same volume. 

These facts give an alternative method for calculating 
the molecular weights of gases, being given the weight 
of a specified volume. As one gram molecule of a gas 
occupies 22.4 liters ^ at standard conditions of temper- 
ature and pressure, the molecular weight of a gas is that 
weight, in grams, which will occupy this volume. To 
calculate the molecular weight of propylene: 298.5 cc. of 
propylene at 18° C. and 768 mm. weigh 0.5315 g. Re- 
ducing this volume to 0° C. and 760 mm., 

It >^ Six 298-5 = 283 cc. 

Then, the molecular weight being the weight of 22.4 liters, 

?|g5 X 0.5315 = 42.07. 

Measurement of Vapor Density. There are three gen- 
eral methods for obtaining vapor density. These are 
the methods of Dumas, Victor Meyer and Hofmann. 

Dumas' Method is applicable to gases and liquids 
which may be vaporized. In this method, a bulb of about 
200 cc. capacity is provided with a narrow neck through 

1 The law of Charles states that gases have the same coefficient 
of thermal expansion. This is not strictly true; although the co- 
efficients are nearly equal, they differ to a slight extent. This 
being the case, all gases at 0° C. and 760 mm. will not occupy ex- 
actly the same volume. 22.4 liters may be taken as an approxi- 
mate value in the majority of cases. This volmne is termed the 
gram molecular volume (G.M.V.). 



GAS CALCULATIONS 97 

which the material is introduced and which is sealed when 
full of the gas to be determined. To make a vapor den- 
sity determination, a small quantity of the liquid to be 
investigated is introduced or the gas is passed in, displac- 
ing the air, after ascertaining the weight of the open bulb 
in air. If used for a liquid the bulb is placed in a water 
or oil bath which is about 25° C. above the boiling point 
of the liquid introduced, which is thereby vaporized and 
escapes through the neck. As soon as no more vapor 
issues from the bulb, the neck is sealed off by a blowpipe 
and the temperature of the bath is taken. After cooling, 
the bulb is again weighed and then the neck is broken 
off under water, which is drawn into the flask, filling it 
completely, except for a small bubble. Filling the bulb 
and weighing the water it holds gives the capacity, the 
barometer reading giving the pressure of the gas. If D is 
the vapor density of the substance, W the weight of the 
vapor in the bulb at the temperature of the bath and 
at atmospheric pressure,^ and W the weight of the air 
which the bulb would hold at the same temperature, then 
the vapor density compared with air is 

W 
D = — • 
W 

The Method of Victor Meyer is not applicable to gases, 
but only to substances which volatilize without decom- 
position, and consists in vaporizing a known weight of 
substance at a temperature considerably above its boil- 
ing point in an apparatus of special form. The vapor 
displaces its own volume of an inert gas, which is col- 
lected (usually over water) and its volume measured, 
noting the temperature and the height of the barometer. 

^ Correction should be made for the buoyant effect of weighing in 
air. This will be treated subsequently. 



98 CHEMICAL CALCULATIONS 

The volume of air is corrected to standard conditions, 
dry. Knowing the weight of substance taken and the 
volume which it occupies in the form of vapor, the vapor, 
the density, or what is more often required, the molecular 
weight, is easily calculated. Suppose 0.1910 g. of sub- 
stance gave 60.50 cc. of air at 11° C. and 752 mm. To 
calculate the molecular weight (vapor density to ^V oxy- 
gen). As the air was measured over water, the volume 
must be calculated to standard conditions, dry. The 
tension of aqueous vapor at 11° C. is 9.81 mm. 

273 ^^ 752 - 9.81 ,, ^^ ^^ ^^ ^^ 
284 ^ — 760 ^ " ^^' 

The weight of vapor which would be required to fill 22.4 
liters (the molecular weight) is 

^^ X 0.1910 = 75.34. 
00.79 

Hofmann's Method is used for substances which may be 
volatihzed undecomposed and consists in vaporizing a 
known weight of substance at a constant known temper- 
ature in vacuo. This is usually done over mercury. A 
graduated glass tube closed at one end is filled with mer- 
cury and inverted in a mercury trough. The height of 
the mercury is noted and a weighed amount of substance 
in a small bottle is introduced into the tube by inserting 
it under the mercury up into the open end of the tube. 
The tube is surrounded by a jacket through which circu- 
lates steam or vapor from some high boiling point sub- 
stance. The temperature maintained in the jacket is 
considerably higher than the boiling point of the sub- 
stance examined. This high temperature drives the 
stopper out of the bottle, vaporizes the substance and 
depresses the column of mercury, the height of which is 



GAS CALCULATIONS 99 

noted.^ From the position of the mercury, the volume 
and the pressure of the vapor inside the tube is deter- 
mined. The temperature of the vapor is read. This 
will be the temperature of the jacket. Suppose 0.0648 g. 
of substance yields 65.30 cc. of vapor at 100° C, the 
barometer reading 752 mm., the mercury in the tube 
standing 480 mm. above the level of the mercury in the 
trough. To calculate the molecular weight of this sub- 
stance (neglecting corrections for the expansion of the 
mercury, its vapor tension and its specific gravity). 

1^ X ^^^75Q^^^ X 65.30 = 17.11 cc. at 0° C. and 760 mm. 

The weight of 22.4 liters, or the molecular weight, is 
22,400 



17.11 



X 0.0648 = 84.86. 



Deviations from the Gas Laws. — Up to the present, 
calculations have been made assuming the absolute 
truth of the laws of Charles and Boyle. These laws 
are only approximations, for, as the critical points of a 
gas are approached or the gas is near a change of state, 
the laws lose accuracy. A gas far removed from its 
liquefaction point acts very closely in accordance with 
the laws of Charles and Boyle; the farther removed, the 
closer its compliance with these laws. It can only be said 
then, that if the gas under consideration acted as an ideal 
gas (an ideal gas is an imaginary gas which obeys the 
laws of Charles and^ Boyle) the predictions would be 

1 By this method, the gas may be obtained under reduced pres- 
sure, which may allow of the examination of substances which de- 
compose on heating to their boiling point under higher pressures. 
Further, the diminishing of the pressure by lowering the boiling 
point allows of a lower temperature being employed for vaporiza- 
tion. 



100 CHEMICAL CALCULATIONS 

correct.^ As an instance, carbon disulphide has a molec- 
ular weight of 76.12, while its specific gravity to ^V oxy- 
gen is 76.4. The volume occupied by a gram molecule 
of this gas is 

76.4 X 0.044656 ^ ^^'^^ ^'*^'"^- 

Also, mercuric chloride having a molecular weight of 
271.52, in gaseous condition shows a specific gravity to 
^V oxygen of 283.5. The volume occupied by a gram 
molecule is then 

271.52 _ 

283.5 X 0.044656 ~ "^^'^^ ^'^^''^• 

This last case is a considerable divergence from the 
theoretical and it will be noted that in this case the vol- 
ume of gas is taken of a substance which at ordinary 
temperatures is a soHd. Water at 0° C. and 76 cm. 
pressure is ordinarily in the solid condition, but at ele- 
vated temperatures becomes gaseous. The same is true 
of mercury and other substances which at ordinary 
temperatures do not exist as gases. It may be predicted, 
however, that if they were in the gaseous state they would 
approximate the volume calculated from the gas equa- 
tions. Thus, if steam, H2O, with a molecular weight of 
18.016 existed as a gas at 76 cm. and 0° C, one gram 
molecule would occupy 22.4 liters. Also, the molecular 
weight of mercury is 200.6 (the atom and the molecule 
of mercury are the same) ; then 200.6 g. of mercury vapor 

1 It is for this reason that some of the answers given to the 
problems do not agree exactly with data obtained from chemists' 
handbooks, it often happening that the specific gravity calculated 
and observed differ slightly. The reason for this is that molecular 
weights are often determined by gravimetric methods which in 
many cases are more accurate than those just described. See 
Chem. Ann., p. 3. 



GAS CALCULATIONS 101 

if it existed as a gas at standard conditions would occupy 
22.4 liters. On this basis it is possible to calculate the 
volume of mercury vapor and the volume of any weight 
of it at any temperature at which it is a gas. One gram 
of mercury vapor at standard (if it were a gas) would 
occupy 

1^X22.4 = 0.1117 liters, 

and at 400° C. the volume of this amount would be 

1^ X 0.1117 = 0.2753 liters, 

and the volume of the vapor at one-half atmospheric 
pressure, 

^ X 0.2753 = 0.5506 liters = 550.6 cc. 

ooU 

To restate the problem: What is the volume of 1.000 g. 
of mercury vapor at 400° C. and under one-half atmos- 
pheric pressure. Solving, by one expression. 

This is a fair approximation to the truth, as the mercury 
would be in the vapor state, the boiling point of mercury 
under atmospheric pressure being 357.33° C. 

Volume of Gas from Weight of Substance (in Metric 
System) . — The foregoing yields a method for calculating 
the volumes of a gas resulting from a given weight of a 
reacting substance. Take the reaction: 

2 KCIO3 = 2 KCl + 3 O2 

(a) Parts by weight (g.) 2 (122.56) = 2 (74.56) + 3 (32) 

(b) Parts by vol.^ 2 vols. 2 vols. 3 vols. 

= 3 (22.4) liters. 
^ Potassium chlorate and potassium chloride are not gases, but 
if they were, the volume 'relations of (6) would hold if they corre- 
sponded to the formulas assigned to them. 



102 CHEMICAL CALCULATIONS 

Volume of Gas from Weight of Substance (in Ounces). 

— By a coincidence approximately the same relations 
exist between the volume of a gas in liters and its weight 
expressed in grams and the volume of a gas expressed in 
cubic feet and its weight expressed in ounces; for one 
cubic foot = 28.32 liters and one ounce = 28.35 g. For 
example, one gram molecule of oxygen has been shown to 
occupy 22.4 liters. To calculate how many cubic feet 
in 32 ounces of oxygen, 

1 oz. = 28.35 g., 1 cu. ft. = 28.32 L.; 
therefore 

32 oz. will occupy 22.4 X 28.35 = 635.04 L., 
and 

635.04 



28.32 



22.42 cu. ft. 



Consequently, gas calculations being rough in most cases, 
it is possible to calculate chemical equations involving 
gases to units in the English system with the same facility 
as with the metric system. Defining an ounce molecule 
as that weight in ounces corresponding to the molecular 
weight, the volume of an ounce molecule of a gas under 
standard conditions is 22.4 cubic feet.^ 

Calculation of Weighings in Air to Values in Vacuo 
and Vice Versa. — The principle of Archimedes as 
stated is that a substance immersed in a liquid is buoyed 
up (loses weight) by an amount equal to the weight of 
the liquid displaced. If for the term ''liquid," the term 
''fluid" is substituted, the law has a more general applica- 
tion, for gases come under the head of fluids. 

Commonly, weighings are carried out in the atmosphere; 
consequently, the weight of an object shown by the ordi- 

1 If the English system is used in all the units, it is better to 
calculate the temperature to Absolute. The other units, such as 
inches of mercury, may be retained. 



GAS CALCULATIONS 103 

nary balance may not be the true weight, as the object 
weighed has displaced a certain amount of air, and con- 
sequently has lost weight by an amount equal to the 
weight of the air displaced. If, however, the weights 
used to counterbalance the object are of the same volume 
they have likewise lost weight to the same extent; conse- 
quently, the mass shown by the counterpoising weights 
is the true mass of the object weighed. If, on the other 
hand, the substance weighed and the weights used to 
counterpoise are of different volumes, the mass may be 
more or less than the true weight. If the volume of the 
substance is greater than that of the weights counter- 
balancing it, the substance has lost weight, and the 
apparent value is smaller than the real value; if the 
weights occupy a greater volume than the object weighed 
they have lost weight to a greater extent than the object 
and the apparent weight of the object is larger than the 
real value. Let W be the true weight in vacuo of the body 

W 

of a density D. The body then occupies a volume -^• 

Let W be the apparent weight of the standard grams 
necessary for counterpoise and D^ the density of the 
standard grams. The volume of the standard grams is 

jy. If c? be the weight of a unit volume of air at the 

W 
time of weighing, then the weights have lost jy d and the 

. W 

loss of weight of the object is yt^- Then the condition 

of equilibrium is 

W W 

W-^d = W'-^,d. 

^ The weights are supposed to be corrected so that in vacuo they 
have this value. 



104 CHEMICAL CALCULATIONS 

Solving for W and W^ gives 
also 

^ D ^ D' 

and since d is small in comparison with D and D\ by 
division the above expressions give 

W = W'+W^d(^^- ^^ ; W = W + Wd (1-, - i). 

The latter form is more convenient for general use as 

TF'(if yr — jyj, or Wdljy — yrj may be solved once for 

all and tabulated for bodies of differing densities as brass 
weights are almost universally used.^ 

Standard (United States) Unit of Volumetric Appa- 
ratus. — The metric system was designed to connect 
the units of length and mass. The gram was intended 
to be the mass of a cube of water, at 4° C, of one centi- 
meter edge. Or, the kilogram was to be the mass of a 
cube of water, at 4° C, of one decimeter edge. This 
relationship was very closely approximated, but the error 
is appreciable, a kilogram of water being found to con- 
tain 1000.027 cc. This discrepancy is very small and 
causes no inconvenience, but because of it, the units of 
capacity for volumetric apparatus have been changed. 
The liter, defined as the volume occupied by a quantity 
of pure water at 4° C, having a mass of one kilogram, and 
the one-thousandth part of the liter, called the milliliter 

1 See Stewart and Gee, " Elementary Practical Physics," Vol. 1, 
p. 90; also Treadwell, "Analytical Chemistry," Vol. 11, p. 14, 1912. 



GAS CALCULATIONS 105 

or cubic centimeter, are employed as units of capacity.^ 
As 4° C. is an inconvenient temperature to work at, 
volumetric apparatus is calibrated to contain, or deliver, 
a specified volume at some other temperature. In this 
country 20° C. has been adopted by the Bureau of Stand- 
ards. At whatever temperature the apparatus is cali- 
brated to contain or deliver, this temperature should be 
marked on the instrument. 

Volumetric instruments are often calibrated by weigh- 
ing into them the calculated amount of water at a definite 
temperature which the instrument should contain, or de- 
liver, to be correct. As the weighing is done in air, cor- 
rection must be made for the buoyant effect of air. The 
water to be weighed may or may not be at the temperature 
at which the instrument is calibrated to be correct, and 
allowance must be made for the expansion of the water 
and the glass. The cubical coefficient of expansion of the 
glass in volumetric apparatus is taken as 0.000025 for each 
degree centigrade. Suppose the temperature at which the 
instrument is calibrated to be correct is 20° C, the expres- 
sion for the volume of water the instrument is to contain, 
or deliver, at other temperatures is 

V = v-\-vX 0.000025 (20 - f ), 

in which V is the volume at 20° C, v the volume at ^° C, 
the temperature (^°) at which the water and the instru- 
ment are at the time of calibrating. 

Thus, to calculate the weight, TF', of water, which should 
be weighed into a flask at ^° C. such that the meniscus 
shall be at the place to be marked to show a correct vol- 
ume at 20° C. Let V be the volume it is desired to have 
the flask contain at 20° C. and d the density of the water 

1 Circular of the Bureau of Standards, No. 9; also Sutton, Vol- 
umetric Analysis, p. 24, 191L 



106 CHEMICAL CALCULATIONS 

at f C. Let h be the buoyancy constant,^ t the temper- 
ature of the flask and water at the time of weighing. Then 
the number of grams W, which should be placed on the 
other pan (the flask being tared by a similar flask and 
beads, foil, etc.), is 

W' = Vd+Vhd-^VX 0.000025 (20 - t), 

in which d is the density of water at the temperature t. 

The Mohr Liter. — Mohr proposed as a unit of vol- 
ume that volume occupied by one gram of water at 17.5° C 
weighed in air with brass weights. In English-speaking 
countries 60° F. or 15.5° C. is commonly used as also is 
15° C. In standard flasks, burettes, etc., the temper- 
ature at which it was calibrated is always marked on 
standard instruments.^ To calculate, for example, the 
true capacity of a Mohr's cubic centimeter: 

Air at standard weighs 0.0012926 g. per cc; then at 
17.5° C, the barometer being at standard: 

^ The buoyancy constant for a body of a specified specific grav- 
ity is obtained by solving ^ ( 7^ " 77/ ) for correcting weighings 

made in air to obtain values in vacuo and conversely ^ ( Ty "" 75 ) for 

converting weights in vacuo to weights in air. This latter quantity 
is intended as the value of (5) in the above. 

2 "Measures marked, e.g., '25 cc. 15° C should contain or de- 
liver, as the case may be, 25 true cc. when, the instrument is at the 
temperature of 15° C. On the other hand, a flask marked ' 1000 g. 
16° C should, of course, contain 1000 g. of distilled water at the 
temperature of 16° C, i.e., a Mohr's liter. Vessels graduated accord- 
ing to Mohr's system should bear the word ' gramme ' or the letters 
'^grm' together with the temperature. It should be noted that the 
German Kaiserliche Normal-Eichungs Kommission no longer em- 
ploys Mohr's unit." Sutton, Volumetric Analysis, p. 24. As to the 
last sentence, the United States Bureau of Standards could also be 
included. 



GAS CALCULATIONS 107 

1 



1 + 0.00367 Xl7:5 >< ''''''''' = '■'''^''' ^' ^'^ ''' 
One gram of water at 17.5° C. = 1.001289 cc, then 
1.001289 X 0.0012146 = 0.0012162 g. air displaced by 
the water. 

^-j = 0.11905 cc. air displaced by the weights. 

0.11905 X 0.0012146 - 0.0001446 g. air displaced by the 
weights. 

0.0012162 - 0.0001446 = 0.0010716 g. loss of weight of the 

water. 

1.0000000 + 0.0010716 = 1.0010716 g. true weight of 1 cc. 
of water under these conditions equals 1.001072 true cc. 

1.001289 - 1.000000 = 0.001289 cc. expansion of the water 
due to change from 4° C. to 17.5° C. 

1.001072 + 0.001289 = 1.00236 true cc. in a cubic centi- 
meter as recommended by Mohr. Then a Mohr's liter 
is 1002.36 true cc, or is 2.36 cc. larger than a true liter. 

Dalton*s Law of Partial Pressures. — The pressure 
exerted by a mixture of gases, or of gases and vapors, or 
vapors, which do not react chemically, is equal to the sum 
of the pressures which each would exert if it alone occupied 
the whole space occupied by the mixture. For simplicity 
a gas saturated with water vapor will be considered. Let 
Pg° = pressure of a dry gas corresponding to a vol- 
ume Vg° and a temperature T°, 
Fg' = pressure of the same gas corresponding to a 
volume V/ and a temperature T\ 

By the laws of Boyle and Charles : 

y o p / mo 

v/ Pg° ^ r ^ [ 



108 CHEMICAL CALCULATIONS 

Let this gas be at F/, P/ and T and let it be saturated 
with water vapor. Let 

y'g+w = volume of the mixture after saturation, 
P'g+w = pressure of the gas after saturation, 
P'w = pressure of aqueous vapor at T\ 

Then, by Dalton's law, 

P^+^. = P'a + P'.; whence P', = P',+, - P',,. 

Substituting in (1) 

V ° P' _ P' TO 

n+. P/ ><r' ^2) 

Now let the gas be at P/, V,° and T°, and let it be sat- 
urated with water vapor. Let 

V°g+y; = volume of the mixture after saturation, 
P°g+w = pressure of the gas after saturation, 
Pw° = pressure of aqueous vapor at T°. 

Then, as before, 

P%+^ = PJ" + P^.^ whence P/ = P%+, - P,°. 

Substituting these values in (2) 

17° . P' jy / rpo 

y/ P° , _ p o >\ 7nJ' 

V g + w J- g + w -Lw J- 

The weight of a moist gas is the weight of the dry gas 
plus the weight of the water vapor. Let 

Wg+w ■■= weight of Vg° volumes of moist gas at a • 
pressure P°g+w and a temperature T°, 
Wg = weight of Vg° volumes of dry gas at a pressure 
Pg° and a temperature T°, 



1 The discussion given on page 85, correcting the volume of a moist 
gas to the volume which the gas would occupy if dry, is a single 
case of Dalton's law. For values of the tension of aqueous vapor, 
see Chem. Ann., pp. 461-466. Mercury also has a small vapor 
tension; see Chem. Ann., p. 468. 



GAS CALCULATIONS 109 

Ww = weight of Vg° volumes of water vapor at a 
pressure Pw° and a temperature T°, 
A = weight of a unit volume of dry gas at Pg° 

and T°, 
B = weight of a unit volume of water vapor at Pg° 
and T\ 
Then 

p/ p f rpo 

W g = p-^ X jpj- X A. 

Since the weight of a gas is directly proportional to its 
pressure and inversely proportional to its temperature, 
and 

p / rpo 

TF„ = ^o X 1^ X B, 
then 

Wg + ^ =Wg+ W^. 

For similar reasons 

p/ p / rpo p f rpo 

rV g + w — p~5 X Jp7 X A -f- -p-5 X ^pT X -D. 

By means of this formula, the weight of a moist gas may 
be calculated for any temperature and pressure. For a 
specified gas, the formula may be considerably shortened. 
Take, for example, air saturated with water vapor. The 
density of aqueous vapor compared to air is 

18.016 ^^00.^ 5 . ^, 

2g^ = 0.62247 = g , approximately. 

Let 

A = weightof aunit volumeof dryairatPp°and jr°, 
B = weight of a unit volume of water vapor at 
Pg° and T°, 
Wa+w = weight of a unit volume of saturated air at 
Pg° and T°. 
Then 

5 = |A. 



110 



CHEMICAL CALCULATIONS 



Then the weight of a unit volume of saturated air at P' 
and r is 

P' 3. P ' TO 

H/a + ^^ p^ X ^ X ^. 

But as the weight of a liter of air is given at 0° C. and 760 
mm., then 

P/ = 760, r = 273 and A = 1.2926 g. 
Substituting, 

TF„^. = ^^^^^^^^^ X P X 1.2926.^ 

Use of this formula is as follows: It is desired to know 
the weight Wa+w of a liter of air saturated with moisture 
at 15° C. {T') under a pressure of 754 mm. (P'^+u,). 
The weight of a liter of a gas is given under standard^con- 
ditions, i.e., temperature 273° A., and pressure 760 mm. 
Of air this weight is 1.2926 g. The tension of aqueous 
vapor PJ at 15° C. is 12.76 mm.^ Substituting in the 
formula, 

„. 273 ^^ 754 - 1 12.76 

^ ^ 288 >< 7^ X 1-2926 = 1.2078 g. 

1 The fraction f is applicable only to air. For any other gas 
the requisite fraction could be worked out in the same manner as 
given for air. 

2 The value 12.76 mm. is given to this degree of accuracy in 
the tables. When subtracted from the barometric reading given in 
millimeters, the result would give a reading which is apparently 
accurate to hundredths of a millimeter, and to be significant, would 
require that the barometric reading was accurate to hundredths of 
a millimeter, which is seldom the case. In the solution of prob- 
lems, however, the vapor pressures have been subtracted from the 
barometric readings as if these were as accurate as the tabulated 
values of the vapor pressures. 



GAS CALCULATIONS 111 

Again, it is required to find the weight of dry oxygen 
contained in a hter of oxygen saturated with moisture 
at 17° C. and under a pressure of 750 mm. (ten. aq. vap. 
at 17° C. = 14.45 mm.). 
97Q 7c;n — 14- 4- Pi 
TF' = 1^ X 760 ^^'^^ 0.044656 = 1.3019 g. O2, 

W" = 1^ X -^ X 18.016 X 0.044656 

= 0.0144 g. H2O vapor, 
W = 1.3019 + 0.0144 = 1.3163 g. 

A gas need not be saturated; a knowledge of the degree 
of saturation will suffice, from which the tension of the 
aqueous vapor may be calculated. In the case of mois- 
ture present in the atmosphere this may be readily deter- 
mined by taking the dew point. The dew point is that 
temperature at which the air would be saturated if it 
contained the amount of water which is present at the 
time and place of determination. For example, it is 
necessary to know the aqueous pressure exerted in the 
laboratory at a certain time. Suppose the dew point is 
found to be 15° C. On consulting the table of aqueous 
tensions ^ the tension of aqueous vapor at 15° C. is found 
to be 12.73 mm. The relative humidity is the ratio be- 
tween the amount of aqueous vapor present in the air at 
a certain time and place, and the amount which would 
need to be present to saturate the air at the temperature 
of the air at the time of taking. As an example: The 
dew point at a certain time and place was found to be 
10° C. The temperature of the atmosphere was 20° C. 
The tension of saturated aqueous vapor at 10° C. is 
9.18 mm.; at 20° C. it is 17.41 mm.; then the relative 
humidity is q -< o 

^ = 0.5273 = 53%. 

1 See Chem. Ann., pp. 461-467. 



112 CHEMICAL CALCULATIONS 

PROBLEMS 

I. 200 cc. of a gas are at a pressure of 752 mm.i and 15° C • 
{a) What is the volume under a pressure of 770 mm tempera* 
ture remaining constant? (6) What is the volume if the tem- 
perature IS lowered to 10° C, pressure remaining constant? 
(c) What IS the volume if changed from 752 mm. and 15° C to 
770 mm. and 10° C? 

Ans. 

752 
(«) ^ X 200 = 195.33 cc. 

(6) 15°C. = 273+15 = 288°A.; 10° C. =273+10 = 283° A. 
2^'X'^^^ = 196.53 cc. 

(^) ^X775X200 = 191.94 cc. 

II. A barometer graduated on a glass scale reads 763.4 mm. 
at 19.5° C. {a) What is the reading corrected to 0° C? (6) If 
the corrected height of a barometer with a brass scale is 764.7 
mm., what does the barometer read at 22° C? (c) If a barom- 
eter with a glass scale reads 754.3 mm. at - 10 C°., what is the 
height corrected to standard? 

Ans. (a) ^^3^ 763.4 . _ . ' 

' ' 1 + 19.5 X 0.00017 - 1.003315 ~ ^^""^ '^"^* 

i + 22 X 0.00016 ^ ^^^'^' 

X = 764.7 X (1 + 22 X 0.00016) = 767.39 mm. 

(c) 754.3 

^^ 1 -10X0.00017 = ^^^-^^^- 

1 Unless otherwise stated, barometer readings in all problems 
are supposed to be corrected to values at 0° C. Barometer readings 
will^ be supposed to be correct as far as the second place past the 
decimal point even if given only to the decimal point. Thermom- 
eter readings will be considered correct to the second decimal place 



i 



GAS CALCULATIONS 113 

ni. A gas at 750 mm. and 12° C, measured moist, has a 
volume of 325 cc. (a) What is its volume, dry, under the 
same conditions? (6) Dry, at standard? (c) 160 cc. of a gas 
are measured moist at 15° C, the barometer (corrected) reads 
743 mm. The mercury in the tube stands 150 mm. above the 
trough; what is the volume of the gas, dry, at standard? 

Ans. 

(a) Tension of aqueous vapor at 12° C. = 10.48 mm. 
'-^^W^' X 325 = If X 325 = 320.46 cc. 

(^)'-^^W^'xix325 = 302.93 cc. 

(c) Tension of aqueous vapor at 15° C. = 12.73 mm. Using 
the decunal coefficient for the change of volume due to change 
in temperature, 1 

743 - (150 + 12.73) ,, 1 

760 >^ r+ 0.00367 X 15 >^^^^=^^^'^''- 

IV. How many cubic centimeters of nitrogen gas, at stand- 
ard, can be obtained from a liter of ammonia gas, at 15° C. and 
780 mm.? 

Ans. 2NH3 = N2 + 3H2, 

2 vol. 1 vol. 3 vol. 

1 27^ 7sn 
2>'288^765>'^^^^ = ^^^-^^^^- 

V. A gaseous mixture contains nitrous oxide (N2O), nitric 
oxide (NO) and nitrogen. Calculate the composition of the 
mixture from the following: 

Volume of mixture taken 51 . 00 cc. 

Volume after adding hydrogen 108. 30 cc. 

1 Where considerable accuracy is desired, the particular decimal 
coefficient of expansion of the gas under consideration should be 
used. See Chem. Ann., p. 73. Unless otherwise stated, the factor 
used in obtaining the answers to problems will be ^^. 



114 CHEMICAL CALCULATIONS 

Volume after explosion 51 . 00 cc. 

Volume after adding oxygen 76 . 50 cc. 

Volume after explosion 57 . 60 cc. 

Ans. Let a; = cc. N2O, 
y = cc. NO, 
z = cc. N2. 
H2 added = 108.3 - 51.00 = 57.30 cc. 
Contraction = 76.50 - 57.60 = 18.90 cc. 
f X 18.90 = 12.60 cc, H2 added in excess. 
57.30 - 12.60 = 44.70 cc. is the required amount of H2 nec- 
essary for the first explosion. The reactions are: 

N2O + H2 = N2 + H2O, 
X -\- X = X -\- 
2 NO + H2 = N2 + H2O, 

^y+y = y+o. 

Then the contraction due to the H2 explosion is 

(1) x + y = 44.70 cc. 

After the first explosion, if no H2 had been added in excess, 
the volume would be z cc. N2 originally present, x cc. N2 from 

the N2O, and | cc. of N2 from the NO; this is 51.00 - 12.60 

= 38.40 cc. Hence 

(2) .T + I + 2 = 38.40 cc. 
Then 

(3) x-{-y+ z = 51.00 cc. (2) 2x + y +2Z = 76.80 cc. 
(2) 2a; H-|/ + 2g = 76.80 cc. (1) x + y = 44.70 cc. 

(4) X + s = 25.80 cc. X + 2^ = 32.10 cc. 

(4) X + g== 25.80 cc. 

^ , . . . 2 = 6.30 cc. 

Substitutmg m (4) gives x = 19.50 cc. 

Substituting these values in (3) gives y = 25.20 cc. 

VI. (a) A liter of sulphur dioxide at standard weighs 2.9266 g. 
Calculate its molecular weight. (6) The molecular weight of 



GAS CALCULATIONS 115 

acetylene is 26.016; what is the weight of 250 cc. of the gas at 

18° C. and 757 mm.? (c) If the specific gravity of hydrogen 
selenide referred to air is 2.806, what is its weight per liter? 
(d) Its molecular weight? 
Ans. 

(«) nnA^f-a =" 65-5; also, 2.9266 X 22.393 = 65.5. 
U.U44O0O 

(6) 26.016 X 0.044656 = 1.1617 g. per L. at standard, 

1^ X 1.1617 = 1.0899 g. per L. at 18° C. and 760 mm. 

7f^7 

^ X 1.0899 = 1.0856 g. per L. at 18° C. and 757 mm. 

^ X 1.0856 = 0.2714 g. m 250 cc. at 18° C. and 757 mm. 
or in one expression, 

26.016 X0.044656xfx§x 11 = 0.2714 g. 

(c) 2.806 X 1.2926 = 3.627 g. per L. 

(d) 2.806 X 28.943 = 81.21. 

Vn. (a) If 30.82 cc. of oxygen (density to air, 1.1055) 
effuses through a small orifice in 55 seconds, what volume of 
hydrogen (density to air, 0.06965) will effuse in the same 
time under the same conditions? (6) What volume of sulphur 
dioxide wiU effuse through a small orifice in the same time as 
83 cc. of ammonia? (c) 150 cc. of air effuses in the same time 
as 63.82 cc. of bromine. What is the molecular weight of the 
bromine? 

Ans. 



a) Substituting in the formula, Fi = V ^ ^ 
V,= \/mMELM^ 122.79 CO. 



0.06965 



116 CHEMICAL CALCULATIONS 

(/;) The densities of sulphur dioxide and ammonia are in the 
ratio of theii' molecular weights; then, as before, 



V, = t/ (83)-X 17.034 ^ 42 go ^^_ 
^ d4.00 

(c) Substituting in the proper formula: 

A = ^{m^ = 5.525, density to air, 
5.525 X 28.943 = 159.89. 

VIII. (a) What volume of oxygen at 18° C. and 754 mm. is 
liberated by 1.763 g. of potassium chlorate when completely 
decomposed? (6) How much sulphuric acid must be taken to 
obtain 5.5 cubic feet of hj^drogen at 17° C. and 762 mm. by 
acting on a metal? 

Ans. (a) 2KCIO3 = 2KC1 + 3 0., 

2 (122.56) g. 3 (22.4) L. 

^•^^^ X^ix^X 3 (22.4)= 0.5193 L. 



2 (122.56) 273 754 
(b) II0SO4 + M" =il/''S04 + H2 

98.09 oz. 22.4 cu. ft. 

IX. (a) A mass of aluminum (density, 2.583) weighed in 
air at 18° C. and 742 mm. showed an apparent weight of 
149.2350 g., brass weights (density, 8.4) being used. What 
is its weight in vacuo? (b) A mass of platinum (density, 
21.48) weighed in an* at 15° C. and 765 mm., against brass 
weights, showed an apparent weight of 89.4130 g. Find its 
weight in vacuo. 

Ans. 

(«) ^f#^xixixo.oo:2.e 

= 0.0684 g. lost by alummum. 



GAS CALCULATIONS 117 

149.2350 273 742 

8.4 ^291 X 760 X U-^Ul^y^^ 

= 0.0210 g. lost by weights. 
0.0684 - 0.0210 = 0.0474 g. difeerence in weights of air 
displaced. 

149.2350 + 0.0740 = 149.2824 g., weight in vacuo. 



(6) Using the formula W =W' + W'd (^ - 4), 

d= i^ X ^ X 0.0012926 = 0.0012333, 

W = 89.4130 + 89.4130 X 0.0012333 f-V. - ^\ 

\21.48 8.4/ 



= 89.4130 + 89.4130 X 0.0012333 (0.04656 - 0.11905), 
= 89.4130 + 89.4130 X 0.0012333 (- 0.07249), 
= 89.4130 - 0.0080 = 89.4050 g. in vacuo. 
X. A flask which is to be marked to contain one liter at 
20° C. (United States Standard) is counterpoised on a balance 
with a similar flask, finishing the counterpoising with sand 
foil, etc. How many grams of water must be weighed into it to 
give the position on the neck which should be marked? The 
conditions are: 

Temperature of air in the balance 22° C. 

Temperature of the water to be used 22° C. 

Temperature of the flask 22° C. 

Barometric pressure 765 mm. 

Saturation of air in the balance 50% 

Vapor pressure of water at 22° C 19. 66 mm. 

Weight of a liter of dry air at 0° C. and 760 mm. 1 . 2926 g. 

Coefficient of expansion of air . 00367 

Density of weights 8.4 

Density of water at 20° C 0. 997797 

Cubical coefficient of expansion of the glass ... . 000025 
Ans. 

Weight of one cubic centimeter of air under these conditions 
is 

1 765 - 0.50 X f X 19.66 

1 + (22 X 0.00367) 760 

X 0.0012926 = 0.0011981 g. 



118 CHEMICAL CALCULATIONS 

The weight of one liter of water in air under these conditions is: 
W = 1000 X 0.997797 + 1000 X 0.997797 

X 0.0011981(3^-^;^), 

= 997.797 + 997.797 X 0.0011981 X (- 0.883159) 
= 997.797 - 1.056, 
= 996.741 g. 
Correcting for the expansion of the glass gives 
1000 X 0.000025 (20 - 22) = - 0.05; 
and the amount to weigh in is 

996.74 - 0.05 = 996.69 g. 

137. One liter of a gas is under a pressure of 780 mm. What 
will be its volume at standard pressure (760 mm.), the tempera- 
ture remaining constant? Ans. 1026.3 cc. 

138. A gas of 300 cc. is under standard pressure. What will 
be its volume at 784 mm., the temperature remaining constant? 

Ans. 290.8 cc. 

139. Five cubic feet of a gas are under a pressure of 27.3" 
of mercury. What is its volume at 29.9", the temperature 
remaining constant? Ans. 4.565 cu. ft. 

140. A gas occupies a volume of 450 cc. under a pressure of 
780 mm. The temperature remaining constant, what pressure 
must be applied to reduce the volume to 400 cc? 

Ans. 877.5 mm. 

141. A gas occupying a volume of one liter, under standard 
pressure, is expanded to 1200 cc. The temperature remaining 
constant, by how many millimeters must the pressure have been 
diminished? Ans. 126.7 mm. 

142. A gas of 200 cc. is at 15.7° C. Find its volume at 0° C. 
(at standard), the pressure remaining constant? 

Ans. 189.12 cc. 

143. One liter of a gas is measured at —15° C. What is its 
volume at 15° C, the pressure remaining constant? 

Ans. 1116.3 cc. 



GAS CALCULATIONS 119 

144. A gas measured 150 cc, at 17,5° C, and on account of 
a change in temperature, the pressure remaining constant, the 
volume decreased to 125 cc. What is the new temperature? 

Ans. - 30.9° C. 

145. The pressure on a confined gas, at 15° C, was 792 mm. 
If the pressure later registered 820 mm., what is the tempera- 
ture then, the volume remaining unchanged? Ans. 25.17° C. 

146. A liter of gas at standard conditions has its tempera- 
ture raised to 15° C. What must be the pressure on the gas, if 
the volume of gas is unaltered? Ans. 801.7 mm. 

147. 183 cc. of a gas are at standard; the pressure is then raised 
to 792 mm. What is the temperature, the volume remaining 
constant? Ans. 11.5° C. 

148. 250 cc. of a gas are at a temperature of 15° C. What 
is the volume of the gas at 0° C, the pressure remaining con- 
stant? Ans. 237 cc. 

149. The pressure on a certain volume of hydrogen is 730 mm. 
at the temperature of melting ice. The volume remaining con- 
stant, what is the temperature at a pressure of 750 mm.? 

Ans. 7.48° C. 

150. Given 250 cc. of a gas under a pressure of 765 mm. and 
at a temperature of 15° C, what is the volume corrected to 
standard conditions (0° C. and 760 mm.)? Ans. 238.5 cc. 

151. A gas measures 300 cc. at standard conditions. What 
is its volume at 755 mm. and 17.5° C? Ans. 321.3 cc. 

152. Under standard conditions of temperature and pressure, 
a gas measures one liter. What is its volume at 17.7° C. and 
748 mm.? Ans. 1082 cc. 

153. A gas of 275 cc. under standard conditions changes its 
volume to 300 cc, when it is under a pressure of 754 mm. What 
is its temperature? Ans. 22.47° C. 

154. 50 cc. of a gas at 780 mm. and at 10° C. changes its 
volume to 48 cc. under a pressure of 792 mm. What is the 
temperature at this pressure and volume? Ans. 2.87° C. 



120 CHEMICAL CALCULATIONS 

155. A gas is at a pressure of 748 mm. and a temperature of 
12° C, when its volume is 200 cc. What must be the pressure 
of the gas if its volume is 178 cc, at a temperature of 0° C? 

Ans. 805.1 mm. 

156. A volume of gas is confined at 0° C. and 760 mm. pres- 
sure, (a) What is this pressure in inches of mercury and (6) in 
pounds per square inch? (Sp. gr. of Hg = 13.6; 1" = 2.54 cm.; 
1 sq. in. = 6.4516 sq. cm.) Ans. (a) 29.92". 

(b) 14.701 lbs. per sq. in. 

157. A barometer with a glass scale reads 752.6 mm. at 
15° C. What is the barometer reading corrected to 0° C? 

Ans. 750.68 mm. 

158. A barometer with a brass scale shows a pressure 768.5 
mm. at 18° C. What is the barometer reading corrected to 
0° C? Ans. 766.28 mm. 

159. The reading of a barometer, with a glass scale, at —5° C. 
is 753.2 mm. What is the reading corrected to 0° C? 

Ans. 753.84 mm. 

160. A brass scale barometer reads 769.2 mm. at 15° F. 
What is the reading corrected to 0° C? Ans. 770.36 mm. 

161. What must be the reading on a barometer with a glass 
scale, at 15° C, that the pressure at 0° C. may indicate 760 mm.? 

Ans. 761.9 mm. 

162. If sufficient water is placed in a dry gas, at 15° C. and 
753.8 mm. to thoroughly saturate it, what would be the pressure 
after saturation, the temperature remaining constant? (Ten. 
aq. vap. at 15° C. = 12.73 mm.) Ans. 766.53 mm. 

163. If the atmosphere is saturated with water vapor at 
14° C. and 758 mm., what percentage, by volume, of water 
vapor does it contain? (Ten. aq. vap. at 14° C. = 11.94 mm.) 

Ans. 1.58%. 

164. A gas, measured moist, has a volume of one liter at 
17.5° C, under a pressure of 758.9 mm. What is its volume, 
dry, under standard conditions? (Ten. aq. vap. at 17.5° C. 
= 14.91 mm.) ^ns. 919.9 cc. 



GAS CALCULATIONS 121 

165. 300 cc. of a gas are measured moist over water at 15° C. 
and under a pressure of 765 mm. (a) What would be its vol- 
ume, dry, at this temperature? (6) What is its volume, dry, 
under standard conditions? (Ten. aq. vap. at 15° C. = 15.38 
mm.) Ans. {a) 295.01 cc. 

(6) 281.48 cc. 

166. A certain reaction produces 22.4 L. of a gas, measured 
at standard conditions, (a) What volume would it occupy, 
moist, at 18° C. and at standard pressure? (6) At 18° C. and 
770 mm.? (Ten. aq. vap. at 18° C. = 15.38 mm.) 

Ans. (a) 24.36 L. 
(b) 24.04 L. 

167. 500 cc. of nitrogen are measured over water at 17° C, 
the barometer reading 750 mm. If the water stood 180 mm. in 
the tube, what is the volume of the nitrogen, dry, at standard? 
(Ten. aq. vap. at 17° C. = 14.91 mm.; sp. gr. Hg = 13.6.) 

Ans. 447.07 cc. 

168. 180.5 cc. of air, saturated with moisture, at 18° C, 
are measured over mercury, the barometer reading 620.3 mm. 
The mercury stood 52 mm. in the tube. Find the volume of 
the air, dry, at standard conditions. (Ten. aq. vap. at 18° C. 
= 15.38 mm.) Ans. 123.7 cc. 

169. 203 cc. of chlorine gas, at standard conditions, are nec- 
essary to decompose a certain amount of hydrobromic acid gas, 
also at standard conditions. What is the volume of hydro- 
bromic acid gas? Ans. 406 cc. 

170. If to a mixture of 100 cc. of nitrogen and of 200 cc. of 
oxygen, 500 cc. of hydrogen are added, and the mixture ex- 
ploded: (a) What is the resultant volume if the water is allowed 
to condense? (6) What is the resultant volume if the water 
stays in the gaseous state? Ans. (a) 200 cc. 

(6) 600 cc. 

171. Consider the following reactions : 

(1) 2 HBr -I- CI2 = 2 HCl + Br2, 

(2) 4 HBr -f- O2 =2 H2O + 2 Bra. 



122 CHEMICAL CALCULATIONS 

(a) If the volume of hydrobromic acid of (1) is one liter, how 
many liters of chlorine at the same conditions are necessary to 
liberate all the bromine? (6) In equation (2) how many vol- 
umes of oxygen of the same temperature and pressure are 
called for to liberate all the bromine? Ans. (a) 0.5 L. 

(6) 0.25 L. 

172. (a) How many cubic centimeters of nitrogen are con- 
tained in a liter of ammonia gas? (b) How many cubic centi- 
meters of hydrogen? Ans. (a) 500 cc. 

(6) 1500 cc. 

173. One liter of hydrogen and 400 cc. of oxygen at 760 mm. 
and 15° C. are exploded in a confined space. What is the vol- 
ume of the resultant gas at 170° C. and 760 mm. pressure? 

Ans. 1538.2 cc. 

174. What is the ratio of the volumes of hydrogen to chlo- 
rine involved when hydrochloric acid is formed? (b) The ratio 
of the volume of chlorine to the volume of hydrochloric acid 
produced? (c) What is the ratio of the sum of the initial 
volumes (factors) to the resultant volume (products)? 

Ans. (a) 1/1. 

(b) 1/2. 

(c) 1/1. 

175. One volume of ammonia gas is decomposed into its 
elements, (a) What is the volume of the resultant gases? 
(6) What is the ratio of the volume of the ammonia gas decom- 
posed to the volume of the nitrogen produced? (c) What is 
the ratio of the volume of the nitrogen to hydrogen produced? 

Ans. (a) 2V. ' 

(b) 2/1. 

(c) 1/3. 

176. 1200 cc. of chlorine and 1500 cc. of hydrogen at the 
same temperature and pressure are combined to form hydro- 
chloric acid gas. (a) What volume of hydrochloric acid gas is 
formed? (6) Which gas is in excess and by how many cubic 



i 



GAS CALCULATIONS 123 

centimeters? (c) What is the total volume of the gases after 
explosion? Ans. (a) 2400 cc. 

(6) 300CC. Ha. 

(c) 2700 cc. 

177. Assuming air to be i oxygen: (a) How many liters of 
air are required to burn one liter of methane at the same tem- 
perature and pressure as the air? (b) If the pressure on the 
methane is two atmospheres and the air is one atmosphere, what 
volume of air is called for? (c) If the pressure on both is the 
same but the temperature of the air is 20° C. and the methane 
0° C. what volume of air is called for? Ans. (a) 10 L. 

(6) 20 L. 
(c) 10.73 L. 

178. 100 cc. of nitrous oxide (N2O) at standard are decomposed 
into its elements and yields 150 cc. of a mixture of nitrogen and 
oxygen. 125 cc. of hydrogen at standard are now introduced 
and the mixture exploded, and the resultant volume found to 
be 125 cc. at standard. Show that the symbol must be N2O. 

179. 300 cc. of hydrogen and 100 cc. of nitrogen both at 
standard are combined to form ammonia gas in a closed system. 
If the volume of the resultant gas is 150 cc. and the pressure is 
unchanged, what is the temperature? Ans. —68.2° C. 

180. 400 cc. of ammonia gas at standard are decomposed and 
the temperature reduced to standard. If the volume of the 
system is unchanged, what is the pressure on the resultant gas? 

Ans. 1520 mm. 

181. 161.53 cc. of hydrogen at 18° C. and 754 mm. are united 
with 160.35 cc. of chlorine at 15° C. and 750 mm. What is the 
volume of the resultant gas at standard? Ans. 300 cc. 

182. A liter of hydrogen at 39° C. is exploded with a liter of 
oxygen at 332.5 mm. The temperature is brought to 0° C. 
What volume of which gas is in excess? Ans. No excess. 

183. 50 cc. of carbon monoxide are combined with oxygen in 
excess. After combination 60 cc. of gas remains. What was 
the volume of oxygen added? Ans. 35 cc. 



124 CHEMICAL CALCULATIONS 

184. 100 cc. of a mixture containing carbon monoxide and 
hydrogen are exploded with an excess of oxygen and 60 cc. of 
gas remain. Of this 60 cc, 40 cc. are absorbed by potassium 
hydroxide (the CO2). What was the volume of the carbon 
monoxide, the hydrogen and the oxygen added? 

Ans. 70 cc. O2, 
40 cc. CO, 
60 cc. H2. 

185. If 50 cc. of ammonia gas at 15° C. and 766 mm. are de- 
composed into its elements, and 50 cc. of oxygen at 12° C. and 
753 mm. are introduced, and the mixture exploded, what vol- 
umes of what gases remain at a temperature of 10° C. and 749 
mm., disregarding the volume of water vapor formed? 

Ans. 25.12 cc. N2, 
12.23 cc. O2. 

186. How many liters of oxygen at 10° C. and 765 mm. might 
be obtained by the decomposition of a liter of nitrogen pent- 
oxide (N2O5) at 35° C. and 748 mm.? Ans. 2.246 L. 

187. Given a mixture of hydrogen, carbon monoxide and 
nitrogen; calculate its percentage composition from the follow- 
ing: 

Volume of mixture taken 25 . 00 cc. 

Volume of oxygen added 10 . 00 cc. 

Volume after explosion 22 . 50 cc. 

Volume after CO2 absorption 12 . 50 cc. 

Ans. 20% H2, 
40% CO, 
40% N2. 

188. Calculate the percentages of hydrogen, carbon monox- 
ide and propane from the following: 

Volume of sample taken 25 . 00 cc. 

Volume after adding oxygen 100.00 cc. 

Volume after explosion 55 . 50 cc. 

Volume after CO2 absorption 17 . 50 cc. 

Ans. 28% H2, 
32% CO, 
40% C3H3. 



GAS CALCULATIONS 125 

189. Calculate the composition of a gaseous mixture of 
carbon monoxide and acetylene from the following data: 

Volume of mixture taken 20 . 00 cc. 

Volume of oxygen added 50 . 00 cc. 

Volume after explosion 52 . 00 cc. 

Volume of oxygen left after KOH treatment. . . 24.00 cc. 

Ans. . 8.00 cc. C2H2, 
12.00 cc. CO. 

190. Calculate the composition of a mixture of carbon mon- 
oxide and methane from the following: 

Volume of mixture taken 20 . 00 cc. 

Volume of oxygen added 40 . 00 cc. 

Volume after explosion 47 . 00 cc. 

Ans. 2.00 cc. CH4, 
18.00 cc. CO. 

191. A gaseous mixture of oxygen, nitrous oxide (N2O) and 
nitrogen is to be analyzed from the following: 

Volume of mixture taken 20 . 00 cc. 

Volume of hydrogen added 30 . 00 cc. 

Volume after explosion 18.00 cc. 

Volume of oxygen added 10 . 00 cc. 

Volume after explosion 19 . 00 cc. 

Ans. 8.00 cc. O2, 
8.00 cc. N2O, 
4.00 cc. N2. 

192. By means of the formulas developed on page 90, cal- 
culate the percentage composition of a mixture of hydrogen, 
carbon monoxide and methane, from the following: 

Volume of mixture taken 50 . 00 cc. 

Volume of oxygen added 75 . 00 cc. 

Volume after explosion 60 . 00 cc. 

Volume after absorption of CO2 20 . 00 cc. 

. Ans. 20% H2, 
40% CO, 
40% CH4. 



126 CHEMICAL CALCULATIONS 

193. What is the weight of 300 cc. of nitric oxide (NO) under 
standard conditions? Ans. 0.40204 g. 

194. What is the weight at standard of a Hter of methane, 
the molecular weight of methane being 16.032? 

Ans. 0.7159 g. 

195. A liter of a gas at standard conditions weighed 3.627 g. 
What is its molecular weight? Ans. 81.22. 

196. 963 cc. of a gas at 754.3 mm. and at 17° C. weigh 1.368 g. 
What is its molecular weight? Ans. 34.05. 

197. Nitrogen has a molecular weight of 28.02. (a) What is 
the weight of a liter at standard conditions? (6) What is its 
specific gravity referred to ^\ oxygen? (c) To hydrogen? 

Ans. (a) 1.251 g. 

(b) 28.02. 

(c) 13.90. 

198. Bromine gas has a specific gravity, referred to air, of 
5.524. What is its molecular weight? Ans. 159.9. 

199. The density of hydrochloric acid referred to air is 1.2595. 
(a) What is its density referred to hydrogen? (b) To oxygen? 

Ans. (a) 18.115. 
(6) 1.1393. 

200. The density of nitrogen at a certain pressure is 0.9802, 
referred to air at standard, (a) What is the pressure upon it? 
(6) What is its density at 15° C. and 760 mm. compared to air 
at standard? (c) If the density compared to air at standard 
is 0.9502, what is the temperature, (d) What is the density 
of nitrogen at 755 mm. and 150° C. compared to air at standard? 

Ans. (a) 769.5 mm. 

(b) 0.9177. 

(c) 5.1° C. 

(d) 0.6207. 

201. A liter of ammonia gas under standard conditions 
weighed 0.7630 g. What is its specific gravity to hydrogen? 

Ans. 8.4885. 



GAS CALCULATIONS 127 

202. A certain volume of chlorine weighs 0.6333 g.; if the 
same volume of silicon tetrafluoride weighs 0.9326 g., what is 
the molecular weight of the latter, taking the molecular weight 
of chlorine as 70.92? Ans. 104.44. 

203. What is the volume at standard conditions of 1.008 g. 
of hydrogen? (b) Of 2.016 g.? Ans. (a) 11.2 L. 

(b) 22.4 L. 

204. The molecular weight of carbon monoxide is 28.00. 
What is its specific gravity to air? Ans. 0.9674. 

205. What is the weight of 275 cc. of carbon oxysulphide 
(COS) at 15.5° C. and 768 mm.? Ans. 0.7054 g. 

206. (a) What is the weight of 300 cc. of chlorine at stand- 
ard conditions if its molecular weight is 70.92? (6) What is 
its specific gravity to A oxygen? (c) To hydrogen? (d) To 
air? (e) To water? Ans. (a) 0.9501 g. 

(6) 70.92. 

(c) 35.24. 

(d) 2.450. 

(e) 0.003167. 

207. A bulb known to hold exactly 500 cc, when filled with 
hydrogen at standard weighed (with all corrections) 30.04494 g. ; 
when filled with ethane under the same conditions it weighed 
30.67105 g. What is the molecular weight of ethane? 

Ans. 30.05. 

208. If the molecular weight of nitrous oxide is 44.02, at 
what pressure would 4 grams of this gas at 15° C. occupy a 
volume of one liter? Ans. 1631 mm. 

209. Calculate the vapor density of carbon tetrachloride, com- 
pared with hydrogen, from the following readings obtained by 
the Hofmann method: 

Weight of substance taken 0. 2832 g. 

Volume of vapor obtained 91 . 50 cc. 

Temperature 99 . 50'' C. 

Barometer 746 . 9 mm. 

Height of mercury in tube 283 . 4 mm. 

Weight of a liter of hydrogen 0.089873 g. 

Ans. 77.05. 



128 CHEMICAL CALCULATIONS 

210. Calculate the molecular weight of an organic substance 
from the following readings obtained by Hofmann's method: 

Weight of substance taken . 0835 g. 

Volume of vapor obtained 81 . 40 cc. 

Temperature 100° C. 

Barometer 752 . 5 mm. 

Height of mercury in tube 413 . 5 mm. 

Ans. 70.38. 

211. What is the vapor density, compared with air as unity, 
of iodine, yielding the following data when determined accord- 
ing to the method of Victor Meyer? 

Weight of iodine taken 0. 1735 g. 

Volume of air displaced 17 . 40 cc. 

Temperature 16° C. 

Barometer 722 . 3 mm. 

Tension of aqueous vapor at 16° C 13 . 57 mm. 

Ans. 8.757. 

212. Victor Meyer's method yields the following results for 
carbon bisulphide. From them calculate the molecular weight. 

Weight of carbon bisulphide taken . 0825 g. 

Volume of air displaced 27 . 34 cc. 

Temperature 16.5° C. 

Barometer 718 mm. 

Tension of aqueous vapor at 16.5° C 14.00 mm. 

Ans. 77.38. 

213. If air has a density of 0.0012926 referred to water: 
(a) What is its density at 754 mm.? (6) At 754 mm. and 38° C? 
(c) At 770 mm. and -40° C? Ans. (a) 0.0012824. 

(b) 0.0011257. 

(c) 0.0015344. 

214. If the barometer at a certain place varied between 
740 mm, and 785 mm. and the temperature varies between 
40° C. and —40° C. : (a) What is the maximum variation in the 
density of the atmosphere if these variations work in conjunc- 
tion? (b) What is the difference in the weight of a liter of air at 



GAS CALCULATIONS 129 

the greatest and the least density? (Density of the atmosphere 
= 0.0012926.) Ans. (a) 0.0004665. 

(6) 0.4665 g. 

215. A certain volume of silicon fluoride weighed 3.4962 g. 
The same volume of air under the same conditions weighed 
0.9694 g. What is the molecular weight of the silicon fluoride? 

Ans. 104.4. 

216. What is the specific gravity to air of hydrogen sulphide, 
the molecular weight of which is 34.086? Ans. 1.178. 

217. The density of a gas is 0.8988 at 15° C. and 749 mm. 
What is its density at standard? Ans. 0.9621. 

218. (a) At what temperature will one gram of hydrogen 
occupy a liter at 760 mm.? (b) At 750 mm.? 

Ans. (a) - 248.42° C. 
(6) - 248.79° C. 

219. 283.5 cc. of a gas at 18° C. and 757 mm. weigh 0.51264 g. 
What is the weight of a liter at standard? Ans. 1.9351 g. 

220. What is the weight at 20° C. and 767 mm. of 263 cc. of 
propylene (CsHg)? Ans. 0.4644 g. 

221. If 283.5 cc. of hydrogen selenide (HsSe) weigh 0.9497 g. 
at 17.4° C. and 747 mm., what is its molecular weight? 

. Ans. 81.19. 

222. If 54.62 cc. of oxygen (density to hydrogen, 15.88) 
pass through a minute opening in the same time as 58.33 cc. of 
nitrogen: (a) What is the density of nitrogen to hydrogen? 
(6) Its molecular weight? Ans. (a) 13.93. 

(b) 28.08. 

223. 60.00 cc. of air effuse in the same time as 53.46 cc. of 
hydrochloric acid. What is the molecular weight of the hydro- 
chloric acid? Ans. 36.45. 

224. 76.42 cc. of sulphur dioxide pass through a minute 
orifice in the same time as 108.1 cc. of oxygen. What is the 
molecular weight of the sulphur dioxide? Ans. 64.03. 

225. It required 75.2 seconds for a certain volume of hydro- 
gen to pass through a minute orifice and 414.6 seconds for the 



130 CHEMICAL CALCULATIONS 

same quantity of carbon oxysulphide. (a) What is the den- 
sity of the carbon oxysulphide to hydrogen? (b) Its molecular 
weight? Ans. (a) 30.4. 

(6) 61.28. 

226. It required 105.6 seconds for a certain volume of air to 
effuse and 39.3 seconds for the same quantity of helium, (a) 
What is the density of helium to air? (b) Its molecular weight? 

Ans. (a) 0.1385. 
(6) 4.009. 

227. It required 87.2 seconds for a certain volume of oxygen 
to pass through a minute orifice and 127.1 seconds for the same 
volume of boron fluoride. What is the molecular weight of 
the boron fluoride? Ans. 67.98. 

228. How many grams of sulphuric acid must act on a metal 
to produce 1200 cc. of hydrogen at 0° C. and 760 mm., all of the 
hydrogen of the acid being liberated? Ans. 5.256 g. 

229. If 15 g. of iron react with sulphuric acid thus: 

Fe + H2SO4 = FeS04 + H2, 
how many liters of hydrogen are liberated at standard? 

Ans. 6.017 L. 

230. In the reaction: 

CaCOs + 2 HCl = CaCl2 + CO2 + H2O. 
(a) How many grams of calcium carbonate are required to pro- 
duce 1700 cc. of carbon dioxide at 780 mm. and 17° C? (b) If 
860 cc. of carbon dioxide were liberated at 754 mm. and 20° C, 
how many grams of calcium carbonate were acted upon? 

Ans. (a) 7.3376 g. 
(b) 3.5515 g. 

231. Lackawanna coal corresponds to the composition: 

Per cent 

Volatile combustible matter 5 . 00 

Carbon 84.00 

Ash 11.00 

Considering the volatile combustible matter to be methane, 
what volume of air at standard, | of which is oxygen, is neces- 
sary to burn completely a kilogram? Ans. 8538.7 L. 



GAS CALCULATIONS 131 

232. Consider the coal of problem just preceding. If the heat 
of combustion of methane burned to carbon dioxide and water 
is 13,344 Calories and of carbon burned to carbon dioxide is 8080 
Calories, what is the heat of combustion of this coal? 

Ans. 7454.4 Cal. 

233. From the equations : 

NaCl + H2SO4 = NaHS04 + HCl 

4 HCl + Mn02 = MnCl2 + 2 H2O + CI2. 

(a) What volume of chlorine measured under standard con- 
ditions will be produced by 10.000 g. of sodium chloride? (b) 
What volume of chlorine measured moist will be produced? 
(Ten. aq. vap. at 0° C. = 4.6 mm.) Ans. (a) 0.9579 L. 

(6) 0.9637 L. 

234. Consider the same reactions as in the previous problem. 
{a) One liter of chlorine measured at 764 mm. (the barometer 
not being corrected for temperature: brass scale) and 15° C, 
moist, indicates how much sodium chloride when the correc- 
tions are made? (6) How much sodium chloride is indicated 
if the readings are assumed to be correct as stated? (c) How 
many grams of sodium chloride are indicated if corrections 
were made for the pressure of water vapor and the tempera- 
ture only? (Pressure of water vapor at 15° C. = 12.73 mm.: 
apparent expansion of Hg = 0.00017.) Ans. (a) 9.757 g. 

(6) 9.948 g. 
(c) 9.782 g. 

235. If ten grams of water vapor occupy a volume of one liter, 
the pressure being 20 atmospheres, what is the temperature? 

Ans. 166.1° C. 

236. Water gas is produced according to the equation: 

C + H2O = CO + H2. 

(a) What volume of carbon monoxide and hydrogen will be 
produced from a kilogram of water? (b) What volume of 
carbon monoxide and hydrogen will be produced from a kilo- 
gram of carbon? (c) What volume of air is necessary for the 



132 CHEMICAL CALCULATIONS 

complete combustion of the gas produced in (a)? (Air = ^ 
oxygen.) Ans. (a) 2486.7 L. 

(b) 3732.2 L. 

(c) 6216.7 L. 

237. A balloon has a capacity of 100,000 liters. How many 
kilograms of (a) sulphuric acid and (b) iron are necessary to 
fill the balloon with hydrogen at 20° C. and 760 mm.? 

Ans. {a) 408 kg. H2SO4. 
(6) 232.3 kg. Fe. 

238. How many liters of air at standard are required for the 
combustion of one kilogram of a coal of the following compo- 
sition? 

Per cent 

Carbon 65.72 

Hydrogen 9 . 03 

Nitrogen 0. 72 

Oxygen 4 . 78 

Ash : 19.75 

Ans. 8475 L. 

239. Given a natural gas of the following composition by 
weight : 

Per cent 

Hydrogen 1 . 42 

Methane 94. 16 

Ethylene 0. 30 

Carbon dioxide . 27 

Carbon monoxide . 55 

Oxygen 0. 30 

Nitrogen 2. 80 

Hydrogen sulphide 0.18 

How much air is necessary for the combustion of one kilogram? 

Ans. 13,598 L. 

240. A sample of calcium carbonate when treated with an 
acid evolved 200 cc. of carbon dioxide at 18° C. and 754 mm. 
How much calcium carbonate does this amount indicate? 

Ans. 0.8316 g. 



GAS CALCULATIONS 133 

241. From the reaction, 

2 Mg + O2 = 2 MgO, 
how many grams of magnesium can be bm-ned in a globe con- 
taining 3489 cc. of oxygen at 19.5° C. and 768 mm.? 

Ans. 7.145 g. 

242. (a) What volume of nitrogen at 10° C. and 624 mm. is 
evolved from 3 liters of nitric oxide at 18° C. and 790 mm. when 
passed over red hot copper? (6) How many grams of copper 
oxide are formed? The reaction is 

2 NO + 2 Cu = 2 CuO + N2. 

Ans. (a) 1846.8 cc. 
(5) 10.392 g. 

243. (a) What volimie of hydrogen at 17° C. and 748 mm. is 
required to unite with one gram of chlorine to form hydro- 
chloric acid? (6) What volume of hydrogen under the same 
conditions but moist? (Ten. aq. vap. at 17° C. = 14.45 mm.) 

Ans. (a) 0.3409 L. 
(b) 0.3476 L. 

244. How many cubic feet of hydrogen at 62° F. and 29.4" 
will be yielded by the action of sulphuric acid on 10 lbs. of 
iron? Ans. 69.475 cu. ft. 

245. 5 cubic feet of carbon dioxide measured at 70° F. and 
30.7" were evolved by the action of hydrochloric acid on cal- 
cium carbonate. How much calcium carbonate was decom- 
posed? Ans. 1.326 lbs. 

246. From the equation, 

NH4NO3 = N2O + 2 H2O, 
(a) how many cubic feet of N2O at standard wiU be produced 
by 15 lbs. of NH4NO3? (b) What is the resultant volume from 
the reaction at 30" and 300° C? Ans. (a) 67.16 cu. ft. 

(6) 422.87 cu. ft. 

247. A gas tank contains 200,000 cubic feet of dry coal gas 
at 30" and 70° F. If 40% by volume of this is methane, how 
many pounds of methane are contained in the tank? 

Ans. 3322 lbs. 



134 CHEMICAL CALCULATIONS 

248. How many cubic feet of oxygen and hydrogen together 
at standard result from the decomposition of | lb. of water? 

Ans. 7.46 cu. ft. 

249. Six cubic feet of sulphur dioxide are required, (a) How 
many ounces of sulphuric acid and (6) copper are required? 
The reaction is 

Cu + 2 H2SO4 = CUSO4 + 2 H2O + SO2. 

Ans. (a) 52.54 oz. 
(b) 17.03 oz. 

250. How many cubic feet of air at standard are required to 
burn a ton of 2000 lbs. of coal gas of the following composition 
by weight? 

Per cent 

Hydrogen 45 . 85 

Methane 39 . 26 

Ethylene 5. 17 

Carbon dioxide . 82 

Carbon monoxide 4 . 78 

Oxygen 0.41 

Nitrogen 3.71 

Ans. 605,515 cu. ft. 

251. If the reaction for the decomposition of oxalic acid 
proceeded as follows: 

2 (C2H2O4 . 2 H2O) = HCOOH + 5 H2O + 2 CO2 + CO, 
and the products were brought to standard conditions, (a) what 
is the volume of the gases (neglecting the volume of water vapor 
present) if 2 oz. of oxalic acid are decomposed? ^ (b) If the 
temperature is now raised to 140° C, what is the volume of 
the gases formed, pressure remaining at standard? (c) If the 
formic acid decomposes at 160° C. as follows: 

HCOOH = CO2 + H2, 
what is the total volume of the gases formed at 200° C? 

Ans. (a) 0.5332 cu. ft. 

(b) 2.420 cu. ft. 

(c) 3.072 cu. ft. 

1 Water freezes at 0° C. and boils at 100° C; formic acid melts at 
8.6° C. and boils at 100.8° C. 



GAS CALCULATIONS 135 

252. A flask weighs 139.8460 g. in air with brass weights 
(density, 8.4). The temperature is 18° C, the barometer 757 
mm. The density of the flask is 3.45. What is the weight 
in vacuo? (Use the decimal coefficient of expansion of a gas. 
1 cc. air at standard weighs 0.0012926 g.) Ans. 139.8748 g. 

253. A flask and water are counterpoised by 1008.2600 g., 
the weights being of brass (density, 8.4). The temperature is 
18° C, the barometer 634 mm. Regarding the flask and water 
as having a density of 1, what is the weight in vacuo? 

Ans. 1009.1586 g. 

254. It is desired to weigh out an amount of platinum (sp. 
gr., 21.5) in air at 17° C. and 764 mm. so that its true weight 
in vacuo should be 200.0000 g. Brass weights are used (den- 
sity, 8.4). What weight should be put on the other pan? (Use 
the decimal coefficient of expansion.) Ans. 200.01774 g. 

255. It is desired to weigh^ out 576.4213 g. to be true in vacuo. 
Brass weights (density, 8.4) are used, the temperature is 19° C, 
the barometer 764 mm. The density of the body to be weighed 
is 2.637; what weight must be put on the other side? (Use the 
decimal coefficient of expansion of a gas.) Ans. 576.2391 g. 

256. It is required to get a weight of water that will be true 
in vacuo. Brass weights are used (density, 8.4), the density of 
the water is 0.999050, the weight required is 100.2037 g., the 
temperature of the air is 18° C. and the barometer is 763 mm. 
What weight must be put on the other pan? (Use the decimal 
coefficient of expansion of a gas.) Ans. 100.0961 g. 

257. A flask of glass (density, 3.4) ^ weighing 203.8050 g. in 
air is filled with water at 17° C. when it weighs 1397.4370 g. in 
air, the temperature being, 17° C, the barometer 756 mm.; brass 
weights are used (density, 8.4). What is the weight in vacuo of 
the combination? (Use the decimal coefficient of expansion of 
a gas. Density of water at 17° C. = 0.9988.) 

Ans. 1398.75 g. 

1 Take into account the weight of air displaced by the glass of 
the flask. 



136 CHEMICAL CALCULATIONS 

258. (a) Calculate ^^ij^f ~ J]j^ for a cubic centimeter of 

water weighed in air under the following conditions: 

Temperature of the air 15° C. 

Temperature of the water 15° C. 

Barometric pressure 770 mm. 

Saturation of aqueous vapor in the air 50% 

Vapor pressure of water vapor at 15° C 12 . 728 mm. 

Coefficient of expansion of air . 00367 

Weight of a hter of dry air at 0° C. and 760 mm. 1 . 2926 g. 

Density of water at 15° C 0. 999126 

Density of weights 8.4 

Coefficient of cubical expansion of glass 0.000025 

(b) Using this last calculated quantity, calculate the appar- 
ent weight, in air, of 250 cc. of water under these conditions, 
(c) If a flask marked 250 cc. at 20° C. (United States Stand- 
ard) gives an apparent weight of water in air under the condi- 
tions enumerated above of 249.508 g., what is the error of the 
flask? Ans. (a) 0.0010903. 

(b) 249.508 g. 

(c) 0.031 cc. 

259. Calculate the weight of a liter of water in air under the 
following conditions: 

Temperature of the air 20° C. 

Temperature of the water 20° C. 

Barometric pressure 760 mm. 

Saturation of aqueous vapor in the air 50% 

Weight of a liter of dry air at 0° C. and 760 mm. 1 . 2926 g. 

Vapor pressure of water at 20° C 17. 406 mm. 

Density of water at 20° C 0. 998230 

Density of weights 8.4 

Coefficient of expansion of air 0.00367 

Ans. 997.173 g. 

260. A flask calibrated to contain 500 cc. at 20° C. (United 
States Standard) was found to hold 499.149 g. of water when 
filled to the mark. The flask was tared with a similar flask, 

1 See p. 104. 



GAS CALCULATIONS 137 

finishing the tare with glass beads, foil, etc. Given the follow- 
ing conditions, what is the error of the flask? 

Temperature of water and flask 17° C. 

Temperature of air 17° C. 

Barometric pressure 760 mm. 

Saturation of aqueous vapor 50% 

Tension of aqueous vapor at 17° C 14.45 mm. 

Weight of a liter of dry air at 0° C. and 760 mm. 1 . 2926 g. 

Coefficient of expansion of air . 00367 

Density of water at 17° C 0. 998801 

Density of weights 8.4 

Cubical coefficient of expansion of glass . 000025 

Ans. +0.15 cc. 

261. A flask is to be calibrated to hold 2000 cc. at 20° C. 
(United States Standard). It is tared by a similar flask, fin- 
ishing the tare with glass beads, foil, etc. How many grams 
of water should be weighed into the flask under the following 
conditions? 

Temperature of water and flask 17° C. 

Temperature of air 17° C. 

Barometric pressure 752 mm. 

Saturation of aqueous vapor in the air 38% 

Vapor tension of water at 17° C 14. 45 mm. 

Coefficient of expansion of air . 00367 

Weight of a Hter of air at 0° C. and 760 mm. 1 . 2926 g. 

Density of water at 17° C 0.998801 

Density of weights 8.4 

Cubical coefficient of expansion of glass . 000025 

Ans. 1995.64 g. 

262. (a) What is the volume of one gram of water at 15° C, 
weighed in air with brass weights (density, 8.4), the tempera- 
ture of the air being 15° C, pressure of the atmosphere 760 mm. 
(b) What is the volume of a kilogram of water under these 
conditions? (Use the decimal coefficient of expansion of a gas. 
Density of water at 15° C. = 0.999126: weight of one cubic 
centimeter of air at standard = 0.0012926 g.) 

Ans. (a) 1.001954 cc. 
(6) 1001.954 cc. 



138 CHEMICAL CALCULATIONS 

263. A flask calibrated to hold one Mohr's liter at 15° C. was 
counterpoised by another similar flask and sand. Filling up to 
the mark with water at 15° C. and 760 mm. required 999.8730 g. 
What is the error? Ans. —0.127 cc. 

264. (a) What is the volume of one gram of water at 60° F., 
weighed in air with brass weights (density, 8.4), the temperature 
of the air being 60° F., the barometer standing at 760 mm. 
(6) What is the weight of a kilogram of water under these con- 
ditions? (Use the decimal coefficient of expansion of a gas. 
Density of water at 60° F. = 0.999050.) 

Ans. (a) 1.002037 cc. 
(b) 1002.037 cc. 

265. If air is saturated with water vapor at 20° C. and 
760 mm.: (a) What per cent of its volume is aqueous vapor? 
(6) Taking air to have a molecular weight of 28.943, what is 
the weight of a liter of saturated air at this temperature? 
(c) What is its density referred to water? (Ten. aq. vap. at 
20° C. = 17.41 mm.) Ans. (a) 2.291%. 

(6) 1.194 g. 
(c) 0.001194. 

266. The temperature of a room is 18° C, the dew point is 
15° C. What is the humidity? (Ten. aq. vap. at 18° C. = 
15.38 mm.; at 15° C. = 12.73 mm.) Ans. 83%. 

267. (a) What is the mass of nitrogen in a liter of nitrogen 
measured moist at 18° C. and 763 mm.? (6) What is the mass 
of the water vapor disseminated throughout the liter? (c) What 
is the mass of the moist nitrogen? (Ten. aq. vap. at 18° C. 
= 15.38 mm.) Ans. (a) 1.1548 g. 

(6) 0.1527 g. 
(c) 1.3075 g. 

268. (a) What is the weight of a liter of water vapor satu- 
rated at 15° C. and 760 mm.? (6) What is its density compared 
to water at 4° C. (Ten. aq. vap. at 15° C. = 12.73 mm.) 

Ans. (a) 0.01277 g. 
(6) 0.00001277 g. 



GAS CALCULATIONS 139 

269. (a) What is the mass of a Uter of saturated air at 15° C. 
and 730 mm.? (6) What is the percentage by weight of air and 
moisture in air under these conditions? (Use the decimal co- 
efficient of expansion of a gas. Ten. aq. vap. at 15° C. = 
12.73 mm.) Ans. (a) 1.1690 g. 

(6) 98.91% ah. 
1.09% H2O. 

270. The humidity at 25° C. is 83%; the barometer stands 
at 765 mm. If the temperature were to suddenly drop to 
15° C. and the barometer to 755 mm., how much moisture would 
be precipitated per cubic meter? (Ten. aq. vap. at 15° C. 
= 12.73 mm.; at 25° C. = 23.55 mm.) Ans. 1.669 g. 

271. How many grams of water will evaporate into a five 
liter space at 20° C. and 760 mm.? (First calculate the density 
of the aqueous vapor at 20° C. to water at 4° C; ten. aq. vap. 
at 20° C. = 17.41 mm.) Ans. 0.0860 g. 

272. (a) What weight of water vapor is present in a cubic 
meter of air saturated at 20° C? (6) If the relative humidity 
is 60%? (c) If the air has a relative humidity of 80% at 
20° C. and should be cooled to 10° C, what weight of water 
would be precipitated per cubic meter? (Density of aqueous 
vapor at 20° C. = 0.0000172; at 10° C. = 0.0000093.) 

Ans. (a) 17.2 g. 
(6) 10.32 g. 
(c) 4.46 g. 

273. What is the weight of a liter of air at 18° C. and 762 mm., 
the dew point being 15° C? (Ten. aq. vap. at 15° C. = 12.73 
mm.) Ans. 1.2050 g. 

274. Calculate the weight of a liter of air 45% saturated with 
water vapor at 765 mm. and 18° C. (Tension of aqueous vapor 
at 18° C. = 15.38 mm. Weight of a liter of dry air at 760 mm. 
and 0° C. = 1.2926 g.) Ans. 1^165 g. 

275. Calculate the weight of a liter of carbon dioxide at 
20° C. and 772 mm., the gas being 50% saturated with water 



140 CHEMICAL CALCULATIONS 

vapor. (Tension of aqueous vapor at 20° C. = 17.41 mm. 
Weight of a liter of dry carbon dioxide at 0° C. and 760 mm. 
= 1.9768 g.) Ans. 1.8585 g. 

276. Using' the symbols of page 109, deduce the approximate 
formula Wg+^„ = ^'''^'' ~^'^^''^ X ^ X 1.9768 for the cal- 
culation of the volume of moist carbon dioxide at any tem- 
perature and pressure to the volume dry at 0° C. and 760 mm. 
(Weight of a liter of dry carbon dioxide at 0° C. and 760 mm. 
= 1.9768 g.) 



CHAPTER VII 

CALCULATION OF ATOMIC WEIGHTS AND 
FORMULAS 

Atomic Weight. — The atomic weight of an element 
is the ratio of the weight of an atom of that element to 
the weight of an atom of some other element taken as 
standard. This standard is oxygen which is arbitrarily 
given an atomic weight of 16. Hydrogen, the lightest of 
the elements, was formerly taken as unity, but this unit 
has been abandoned and oxygen has supplanted it, for 
the reason that most of the elements form binary and 
other compounds with oxygen which are stable at ordi- 
nary temperatures, while the same is not true of hydrogen. 
Again, it happens that when oxygen is taken as 16, more 
of the atomic weights of the elements can be expressed in 
whole numbers than is the case when the hydrogen stand- 
ard is employed, and also there are more compounds 
of importance to the chemist, containing oxygen, than 
hydrogen. 

Valence. — Valence is the power of an atom of an 
element to combine with a certain number of atoms of 
another element, taking the valence of hydrogen as unity. 
Valence is a numerical property of an element by virtue 
of which a definite number of atoms of another element 
are held in combination. 

Combining Weight. — The combining weight of an 
element is the weight of that element which will combine 
with, replace or be replaced by, 16 parts by weight of 
oxygen. 

141 



142 CHEMICAL CALCULATIONS 

Radicals or ions may be considered to have valence. 
Thus, sulphuric acid has two replaceable hydrogens, so 
the radical SO4 is divalent: the hydroxyl radical, OH, 
combines with one atom of hydrogen to form water, one 
atom of sodium to form sodium hydroxide and, conse- 
quently, has a valence of one. 

Relationship between Atomic Weight, Valence and 
Combining Weight. — As oxygen has a valence of two, 
the following equation shows the relationship between 
the atomic weight, the combining weight and the valence 
of elements 

A = -V 

in which A is the atomic weight, C the combining weight ^ 
and V the valence. 

Calculation of Atomic Weight from an Analysis. 
Simple Cases. — For example, mercury and oxygen 
have each a valence of two, the compound mercuric oxide 
is HgO, which by analysis gives: 92.61% mercury and 
7.39% oxygen. Then X grams of mercuric oxide contain 
0.926 X g. of mercury and 0.074 X g. of oxygen. The 
equivalent of mercury according to definition is 

0.9261 X ,,.,^ 09261 ^_ ^„ „ . 
0:0739X >< 1^ = 00739 >< 1^ = 200.6. 

* Combining weights may be referred to other elements than 

oxygen. If x represents the valence of the element to which the 

C 
combining weight is referred, then the formula becomes A = — V. 

X 

The term "equivalent" is often employed and by it is meant that 
weight of an element which will combine with, replace or be re- 
placed by 8 parts by weight of oxygen. With this definition the 
formula becomes A = CV. This latter is frequently used, but the 
formula given in the text is employed for the reason that it refers to 
oxygen as 16, which is the arbitrary standard to which all atomic and 
molecular weights are referred. 



ATOMIC WEIGHTS AND FORMULAS 143 

Substituting in the equation, the atomic weight is 
A = ?f:6 X 2 = 200.6. 

Again, iron in its trivalent form is combined with oxy- 
gen yielding an oxide which on analysis shows 30.06% 
oxygen, 69.94% iron Then, as before, the equivalent of 
the iron is 

0-6994 

0.3006 ^ ^^ " ^^'^^' 

and the atomic weight of iron is 

A = ?^ X 3 = 55.84. 

C 
Law of Dulong and Petit. — The formula A= ^ V^ 

supposes a knowledge of the valence of the element the 
atomic weight of which is to be calculated. The com- 
bining weight is found by an analysis of a compound, 
preferably a compound of the element with oxygen, if 
such compound be stable. In the case of compounds 
of elements which cannot be vaporized without decom- 
position, a knowledge of the valence of the element must 
be determined by other means than through molecular 
weights determined by vapor density methods. For this 
the law of Dulong and Petit is employed. This law 
states that the product of the atomic weight of an element 
into its specific heat (in the solid state) is a constant. 
The product is called the atomic heat and its value approx- 
imates 6.4.2 jf yj represents the atomic weight and S 
the specific heat, the law may be symbolized: 
Atomic heat = Sw = 6.4. 

1 P. 142. 

2 Exceptions occur, notably in the cases of carbon, silicon, beryl- 
lium and boron. 



144 CHEMICAL CALCULATIONS ' 

Values of the atomic weights so obtained are rough 
approximations, but they serve to indicate which of a 
series of values must be taken as the atomic weight. 
This is shown in the following: The specific heat of iron 
is 0.1162;^ and analysis of ferric oxide gives 

Per cent 

Iron 69.94 

Oxygen 30.06 

100.00 

The problem is to find the atomic weight and the valence 
of iron in this compound. The combining weight of iron 
from these figures is 

3^^X16.00 = 37.227. 

Then as oxygen has a valence of two, the lowest possible 
value for the atomic weight of iron is half the combining 
weight, or 18.6135; Fe20 being a compound in which 
iron would show a minimum atomic weight as then its 
valence would be one. Without taking into consider- 
ation any other compounds of iron, the atomic weight 
of iron might be any whole multiple of this value, namely, 
18.6132, 37.227, 55.8405, 74.454, etc. But from the law 
of Dulong and Petit the atomic weight of iron must 
approximate 

'"=01^2 = ^5.08. 



This shows 55.84 to be the atomic weight of iron. From 
the form 
must be 



C 

the formula A = -V, the valence of iron in ferric oxide 



55.84 = ?^F; Y=Z. 
* See Chem. Ann., p. 6. 



ATOMIC WEIGHTS AND FORMULAS 145 

Calculation of Atomic Weights from Analysis. More 
Complex Cases. — The above are examples of the 
simplest cases. A more complex example will now be 
considered. Arsenious oxide (AS2O3) is converted into 
silver arsenate (Ag3As04): 1.5000 g. of arsenious oxide 
yield 7.0096 g. of silver arsenate. Knowing the symbols 
and the atomic weights of all the elements except the 
atomic weight of arsenic, which is to be determined, 
the conversion may be represented 

AS2O3 — > 2Ag3As04 

(a) 1.5000 g. 7.0096 g. 

(6) 48 + 2 X 775.28 + 2 X 

The subscripts (a) show the ratios by weight, (6) the 
ratios by molecular weights in which X is the atomic 
weight of the arsenic. These two ratios are equivalent: 
1.5000 ^ 48 + 2X 
7.0096 "775.28 + 2 X' 
and solving for X gives X = 75.00 as the atomic weight 
of arsenic. 

Again, consider the equation, 

NaCl + AgNOg = AgCl + NaNOg, 

and suppose it were determined that 10.0000 g. of sodium 
chloride require 18.4535 g. of silver dissolved in nitric acid 
for complete precipitation. To calculate the atomic 
weight of sodium: According to the statement, 

10.0000 g.NaCl = 18.4535 g. Ag, 
the amount of chlorine in the silver chloride is the amount 
of chlorine in the sodium chloride, which is 

S -Sfix 18.4535 = 6.0657 g.Cl, 

and the weight of sodium in the sodium chloride is 
10.0000 - 6.0657 = 3.9343 g. Na; 



146 CHEMICAL CALCULATIONS 

consequently, the atomic weight of sodium is 
3.9343 



6.0657 



X 35.46 = 23.00. 



The molecular weight of a substance is the sum of the 
atomic weights of the atoms entering into the molecule. 
When a substance may be vaporized without decom- 
position, the molecular weight is obtained as given in 
the section devoted to gas densities. 

Molecular Weight by Elevation of Boiling Point and 
Depression of Freezing Point. — In addition to the 
method of measuring vapor densities, there are two 
methods commonly employed with non-electrolytes (i.e., 
substances which do not ionize appreciably upon going 
into solution). These may be determined, in general, 
by measuring the depression of the freezing point and 
the elevation of the boiling point, due to the introduction 
of a solute. 

The depression of the freezing point of a given amount 
of solvent is the same for a gram-molecule of substance, 
and, in general, is independent of the composition of the 
substance if no ionization takes place. The amount of this 
depression depends upon the nature of the solvent. Let 
M be the molecular weight of the dissolved substance, co 
its weight in grams, W the weight of the solvent in grams, 
A the observed change of temperature and K a constant 
depending upon the nature of the solvent. The molecu- 
lar weight is given by the formula 

Using the symbols with the same meaning, as above, 
for the determination of molecular weights from the 
raising of the boiling point, the same formula applies 
except that K is a different constant and A represents 



ATOMIC WEIGHTS AND FORMULAS 



147 



the elevation of the boiHng pomt of the solvent due to 
the introduction of the solute. Below are a few values 
ioiK: 



Solvent. 


K 
Dep. of F. P. 


K 
Elev. of B. P. 


Water 

Benzol 

Ethyl ether 


1850 
5000 


510 
2610 
2160 
3590 


Chloroform 









Molecular Weight of an Organic Acid from Analysis 
of the Silver Salt. — A method for the determination 
of molecular weights particularly applicable to organic 
acids is illustrated as follows : ^ One gram of silver ace- 
tate on analysis is found to contain 0.6464 g. of silver. 
As silver is univalent, then one hydrogen must have been 
replaced for each acidic hydrogen ion in the acetic acid, 
which being a monobasic acid has one replaceable hydro- 
gen ion. It is evident that in this compound there are 
1.0000 - 0.6464 = 0.3536 g. 

of substance other than silver; the molecular weight of 
the radical is 

0.3536 



0.6464 



X 107.88 = 59.015. 



The weight of the acetic acid, then, must be 
59.015 + 1.008 = 60.02. 

Molecular Weight of an Organic Base from Analysis 
of the Hydrochloro Platinic Acid Salt. — The molecular 
weight of an organic base may be determined by weigh- 

^ Strictly speaking, this method presupposes a knowledge of the 
number of replaceable hydrogen atoms and in reality only gives the 
molecular weight of the empirical formula of the acid. 



148 CHEMICAL CALCULATIONS 

ing the hydrochloro platinic acid salt of the base, igniting, 
and weighing the residue of platinum after ignition.^ 
The formation of the salt may be symbolized (letting B 
represent the organic base) as follows: 

2 B + HsPtCle = BaHsPtCle. 
One gram of methyl amine hydrochloro platinic acid 
leaves 0.4134 g. of platinum after ignition. To cal- 
culate the molecular weight of methyl amine, let x 
represent the molecular weight of methyl amine, then 
1.0000 ^ 2x + HgPtCls ^ 2x-\- 409.98 
0.4134 Pt 195.2 

Solving gives x = 31.08. 

Formula of Compound, Given Molecular Weight and 
Percentage Composition. — The molecular formula of 
a compound is easily determined, knowing the molecular 
weight and the percentage composition. Take, for exam- 
ple, barium dichromate, given the following: 

Molecular weight 353. 57 

Barium 38.85% 

Chromium 29.47% 

Oxygen 31.68% 

The proportion of barium in the molecule is (atomic 
weight = 137.37) 

105:06 ^ ^^^'^^ ^ ^^^-^^^ 137^37 " ^'' 
of chromium (atomic weight = 52.1), 

^^■'^'^ X 353 57 -104 2 ^^■^-2- 

ioaoo ^ ^^-^-^^ " ^"*-^' kT ~ ^' 

of oxygen (atomic weight = 16.00), 

^„ X 353.57 = 112, f = 7. 

Consequently, the symbol is BaCr207. 

1 This method presupposes the organic base to be mono acidic. 



ATOMIC WEIGHTS AND FORMULAS 149 

Formula from Percentage Composition. — Formulas 
may be empirical or molecular. The empirical formula is 
the simplest formula that can be assigned the substance 
and, in general, is only used in the absence of knowledge 
of the molecular formula. The molecular formula shows 
the number of atoms in the molecule and consequently in- 
dicates the molecular weight, that is, that weight of sub- 
stance which in the gaseous state would occupy 22.4 liters. 
To calculate the empirical formula of a compound, let 

A = symbol of one of the elements present in a com- 
pound ; 

B = symbol of another element present in the same 
compound ; 

C = symbol of still another element present in the same 
compound. 

X = subscript of A in the compound under consideration; 

y = subscript of B in the compound under consideration; 

z = subscript of C in the compound under consideration. 

a = atomic weight of the element A ; 

h = atomic weight of the element B; 

c = atomic weight of the element C. 
AxByCz = molecular symbol of the compound under con- 
sideration. 

Then the molecular weight of the compound is ax-\-by-\- cz. 
Let this be represented by M. Then 

%A=^X100; %B=gxlOO; % C = g X 100. 

Then 

%A : %B : %C = ^X 100 : §X 100 : f X 100, 

= ax :hy : cz. 
Hence 



150 CHEMICAL CALCULATIONS 

Then 

in which X is a constant. Multiples of K are chosen 

which will give x, y and z integral values. 

To apply the above, suppose a substance analyzes 

Per cent 

Carbon 40.0 

Hydrogen 6.6 

Oxygen 53 . 4 

Then if x, y and z represent the subscripts of carbon, 
hydrogen and oxygen respectively, 

-^ = 3.3 

. = ^ = 3.3 

As there cannot be less than one atom of each element 
present, it is evident that the ratio of the atomic weights 
of carbon, hydrogen and oxygen in this compound must 
be one atomic weight of carbon, one atomic weight of 
oxygen and two atomic weights of hydrogen, and the 
empirical or simplest formula is CH2O, corresponding to 
a molecular weight of 30.^ Such a compound is known 
and is formaldehyde (oxymethylene) . The compound 
might, however, be (CH20)x, and in the absence of knowl- 
edge of the value of x no further information may be 

1 The value of K in this example is -^-z . Multiplying the values 

0,0 

obtained for x, y and z by this quantity gives integral values for x, 
y and 2; of 1, 2 and 1 respectively. A value for K need not be de- 
termined as will be seen later. 

It is not necessary to have the composition of the compound 
expressed in percentages. For instance, the same result is obtained 



ATOMIC "WEIGHTS AND FORMULAS 151 

conveyed than is given in the empirical formula. There 
is a compound yielding the same percentage composition 
which shows a molecular weight of 60 and corresponds 
to the molecular symbol C2H4O2 (acetic acid). In this 
case the value of x is two. Another compound, which on 
analysis has the same composition and has a molecular 
weight of 90, corresponds to the formula CsHeOa and is 
lactic acid. Finally, there is still another compound of 
molecular weight 180, being grape sugar, C6H12O6, in 
which the value of x is six. When the molecular weight 
of a compound is known the molecular formula may be 
calculated: when this is lacking, given the percentage 
composition of a compound, in the absence of other knowl- 
edge, the empiric formula only may be determined. A 
complex case will now be considered. A compound 
analyzes : 

Per cent 

Carbon 49.97 

Hydrogen 6 . 46 

Nitrogen 17.95 

Oxygen 25.62 

100.00 



from the following analysis of this same compound, 2.6000 g. of 
substance being taken: 

Carbon 1 .0400 g. 

Hydrogen . 1716 g. 

Carbon 1.3884 g. 

2.6000 g. 
Then 

' C =^ = 0.08667 = 1, 

O = ^^ = 0.08679 = 1, 
as before. 



152 CHEMICAL CALCULATIONS 

To calculate its empiric formula. As before, 
. = ^^2^4.164, 

_ 17.95 _ 
^ - 14:01 - ^■^^^' 

= 2462^1.601. 
16 

The compound, then, corresponds to the formula C4.164 
H6.406Ni.28iOi.6oi. But a formula must express the ratio 
of the atoms in simple whole numbers, the problem now 
presented being to find this ratio. It is apparent that 
the nitrogen atoms are present in the smallest number; 
then taking nitrogen as unity as in (a) : 

(a) (6) 

4 164 4 164 

c = mr = 3-2^0' 06405 = 6-^oi 

6.406 6.406 

^ = 1:281 ^-^ 06405 ^°°0 

N = l-S - 1-000 0W5 = 2-000 

1.601 1.601 

^ = 1:281 = 1-250 0:6504 = 2-500 

^ The ratios in (a) assume nitrogen to be present to the extent 
of one atomic weight. With this assumption it is seen that the 
other elements are not present as whole but as fractional atomic 
weights, which, according to the atomic theory, cannot be. The 
ratios of (6), (c) and {d) may be arrived at by multiplying these 
ratios by 2, 3 and 4 respectively. Finally, a series of ratios ap- 
proximating whole numbers is reached. 



ATOMIC WEIGHTS AND FORMULAS 153 

(c) (d) 

4 1 64 4 1 64 

From (a) the formula would be C3.25H5N1O1.25. This 
ratio is not in whole numbers. Determining the ratio 
on the assumption that two nitrogens are present (by 
dividing the nitrogen ratio, 1.281, by 2, giving the quotient 
0.6405 and dividing through by this), the formula C6.5H10 
N2O2.5 is obtained (6). Taking nitrogen as being present 
to the extent of three atoms, the ratio as found in (c) is 
C9.75H15N3O3.75. Assuming four nitrogen atoms to be 
present, the ratio numbers of (d) yield C13H20N4O5. The 
case cited has been carried to the end and this method 
must yield results finally, but only on the assumption 
that the percentage composition as given is correct. 
That this ratio would finally have been reached might 
have been seen by an inspection of the ratios of (a). 
These ratios (a), it will be observed, are all evenly divisible 
by 0.25; consequently, dividing the ratios through by 
this number would have led to the result, C13H20N4O5, by 
inspection, obviating the necessity of calculating (b), (c) 
and {d)} Such a short cut is not always apparent at a 

1 The process amounts to finding the approximate lowest com- 
mon multiple of the subscripts. Knowing the lowest common 
multiple, by dividing the subscripts by it, the numbers so obtained 



154 CHEMICAL CALCULATIONS 

glance, in which event the operations must be continued 
till approximately whole numbers are obtained.^ 

Most acids, bases and salts may be regarded as being 
composed of two oxides.^ Na2S04 may be considered as 
made up of the basic oxide Na20 and the acidic oxide 
SO3; as Na20 • SO3. The analyses of many substances 
report the constituents present as oxides; particularly is 
this so in mineral analysis. These radicals or oxides may 
be treated in the same way as elements. A substance 
shows: 

Per cent 

Potassium oxide (K2O) 9 . 93 

Aluminum oxide (AI2O3) 10. 77 

Sulphur trioxide (SO3) 33.72 

Hydrogen oxide (H2O) 45 . 56 

yield the subscripts desired. As analyses contain unavoidable 
errors, the true mathematical lowest common multiple will not 
serve, hence it is best to proceed by some such method as outlined. 
Just where to stop depends upon the degree of accuracy of the 
analysis. 

1 The slide rule is of great assistance in calculations of this 
nature, as when set to show a ratio between two numbers, all num- 
bers coinciding are in the same ratio. 

2 Halogen salts, sulphides, and general salts of acids which do 
not contain oxygen, may not be so considered. When an analysis 
shows acidic and basic oxides and also a halogen or halogens re- 
ported as such, the analysis, no matter how carefully done, even if 
theoretically perfect, will total more than one hundred per cent. 
This is for the reason that the metal combined with the halogen is 
calculated to the oxide where in fact it was present as a halogen salt. 
Hence, if the metal is calculated to an oxide and the halogen reported 
as such, the analysis will show too great a total by an amount equal 
to the oxygen calculated to be present but which was not present. 
Such an analysis will be too high by an amount equal to "the oxy- 
gen value of the halogen" or ^-Ft X % M, in which H is the molec- 
ular weight of the halogen and % M represents the per cent of the 
halogen reported. 



ATOMIC WEIGHTS AND FORMULAS 155 



The calculation of its formula is 




K.0 = g| =^ 0.1054 : 


0.1054 
0.1054 ^' 


AI2O3 = Jg = 0.1054 : 


0.1054 
0.1054 ' 


^0^ = Urn = '-'''' ' 


0.4211 
0.1054 ' 


H.O=f^ = 2.529 : 


2.529 _ 



and the compound is K2O • AI2O3 • 4 SO3 • 24 H2O, or 
it may be written: K2O • SO3 : AI2O3 • 3 SO3 : 24 H2O 
AI2 (804)3 • K2SO4 • 24 H2O, or KAl (804)2 • 12 H2O. 

In the problem above, the analysis is supposed to be 
correct as given. In practice, the analysis of a substance 
will contain errors, large or small, according to the accuracy 
of the methods employed and the difficulties encountered 
in the determination of the various constituents. Where 
the constituents are present in about equal amounts and 
all the constituents are determined with the same degree 
of accuracy, it is easier to find the ratios of the atoms or 
radicals by the method already given. In many cases, 
one or more constituents are present to but a small extent, 
or the methods by which they were determined do not 
lend themselves to the same degree of nicety of deter- 
mination as with some of the other constituents present. 
In the calculation of formulas many things must be taken 
into account and it is largely a matter of judgment as to 
the manner of calculation. These facts, the method of 
preparation or the manner of formation of the compound, 
its similarity to other well-known compounds of the same 
type, the methods employed in the analysis: all these 
must be taken into account, leaving the determination 



156 CHEMICAL CALCULATIONS 

of the formula of a compound to a large degree a matter 
of judgment. Suppose a substance on analysis yields 

Per cent 

Manganese 24.50 

Hydrogen ^ 1 .02 

Arsenic 45 . 07 

Oxygen (by diff.) 29. 41 

and let it be assumed that the arsenic is determined with 
the greatest degree of accuracy. Then 

1 02 
H = 008 = 0-902. 

^^ = ftS = 0-602, 
= 29^^,838, 



and the compound approximates Mn0.446As0.601O1.838H0.902. 
It is required to find the ratio of simple whole numbers 
which approximates these ratios. By inspection it will 
be noted that the ratio of manganese to hydrogen is 
approximately 1:2. Also it can be seen that the ratio 
of hydrogen to arsenic is nearly 3 : 2. Then it is evident 
that if these numbers representing the ratios found by. 
analysis are divided by 0.15 that the ratios of manganese, 
hydrogen and arsenic fall into approximately even num- 
bers. This would indicate the arsenic to be present to 

the extent of four atoms (^j^ "" ^)* Then, proceeding 
on this assumption: 



ATOMIC WEIGHTS AND FORMULAS 157 

Mn 



0.446 

^ ^ - = 2.97 = 3 very closely, 
U.io 



H = ^ = 6.01 = 6 very closely, 

. 0.602 . ^, . , 1 

As = = 4.01 = 4 very closely, 

U.lo 

1 CQO 

O = ^^^ = 12.2 = 12 very closely. 
U.io 

Consequently, the compound is Mn3H6As40i2 or MnaHe 

(AS03)4. 

Formulas of Minerals. Isomorphic Replacement. — 

A definite crystalline structure generally indicates purity 
and, consequently, definite chemical composition. Mit- 
scherlich propounded the law that: ''Substances which 
are analogous chemical compounds have the same crystal- 
line form"; in other words, they are isomorphous. To a 
large extent this is true, but there are so many exceptions 
that the statement cannot be called a law. The ''alums" 
consist of sulphates of a trivalent and a monovalent base 
with twelve molecules of water, all crystallizing isomor- 
phously. By manipulation a crystal may be grown to 
contain one or more alums. Such crystals will not show 
definite chemical composition, but are mixtures, and may 
be regarded as a single alum, parts of which have been 
replaced by an analogous substance. This is isomorphous 
replacement. 

Strict rules for isomorphous replacement cannot be 
given. In general, it may be said that metals and metallic 
oxides of the same structure can replace each other and 
the same is true of the non-metals and acidic oxides. 
Thus AI2O3 and Fe203 are mutually replaceable as are the 
oxides FeO, MgO, CaO, MnO, etc., these latter being 
similar in being oxides of divalent metals. In like manner 



158 CHEMICAL CALCULATIONS 

the alkalies, NH4, Li, Na, K, etc., and their oxides (NH4)20, 
Li20, Na20, K2O, etc., may replace each other in a mineral. 
This is not all — CaO, for example, may be replaced by 
K2O or another alkali oxide, as also, in some instances, FeO 
may be replaced by AI2O3. This leaves the matter to a 
large extent one of judgment: for more complete informa- 
tion the student is referred to works on mineralogy. The 
more common cases follow the rule of compounds of like 
structure replacing each other. 

Mosander examined a specimen of Menaccanite and 
found it to be composed of 

Per cent 

Ti02 46.92 

FeaOs 10.74 

FeO 37.86 

MnO 2.73 

MgO 1.14 

99.39 

The percentages of the metals and oxygen are 
m ^ £1 ^ 46.92 = 28.18% Ti and 18.74% 0, 

^ = g|^ X 37.86 = 29.44% Fe and 8.42% O, 
^ = fJII X 2.73 = 2.12% Mn and 0.61% O, 

my = SS>< ^-^^ = ^-^^^^^ and^5%0, 

Total 31.45% 0. 

^ These calculations are very easily performed by the use of the 
slide rule, as the error of the rule is, on the average, less than the 
error of analysis, etc. For the same reason approximate atomic 
weights may be used. 



ATOMIC WEIGHTS AND FORMULAS 159 

If the metals replaced iron, the percentages replaced are 
Fe 55.85 



Ti 48.1 
j;e ^ 55.85 
Mn 54.93 

Fe ^ 55.85 
Mg 24.32 



X 28.18 = 32.71% Ti as Fe, 
X 2.12= 2.16%MnasFe, 
X 0.68= 1.58%MgasFe, 



36.45 + 36.95 = 73.40% total Fe. 
T, 73.40 , ^,, 1.314 ^^^ 
^^ = 5-5:85 = 1-^1^ '• 0-:657 = ^'^^^ 

'' = 'ifo= '-''''' S=2.99 = 3approx. 

This mineral, considered as an oxide of iron,^ approximates 
very closely to the formula Fe203 (hematite), part of the 
iron being replaced by other metals, chiefly titanium, 
making it (TiFe)203. The mineral as it analyzes is 
(Ti, Fe, Mn, Mg)203. An inspection of the analysis 
would show that there are no distinctively acid oxides 
present, which would suggest the method of calculation 
adopted. 

^ The metals could just as well be calculated on the supposition 
that they replaced titanium: 

~ = ^i X 36.95 = 31.82% Fe as Ti, 
J^e 00. oo 

Ti 48.1^ 2.12= 1.86%MnasTi, 



Mn 54.93 
Ti ^ 48.1 
M.g 24.32 



X 0.69 = 1.37% Mg as Ti, 



35.05 + 28.18 = 63.23% total Ti. 
Ti = ' = 1.315 = 2 approximately, 

O = .,' = 1.966 = 3 approximately, 
lb 

and the mineral becomes Ti203, or more generally M2O3, as before. 



160 CHEMICAL CALCULATIONS 

Given the following analysis by Pisani, to determine the 
formula of the mineral. 

Per cent 

SiOz 45.95 

AI2O3 0.85 

FeO 8.91 

MnO 10.20 

ZnO 10.15 

CaO 21.55 

MgO 3.61 

Loss on ignition 0.31 

101.53 

This mineral is obviously a silicate. As a silicate may 
be regarded as a combination of basic and acidic oxides, 
the calcium oxide being present in the greatest amount, 
the metals will be calculated to calcium oxide. Disre- 
garding the loss on ignition: 

fS = ?iS ^ 8.91 = 6.96% CaO - FeO, 

^ = f|§-X 10.20 = 8.07% CaO - MnO, 

^ = |f§ X 10.15 = 7.00% CaO ^ ZnO, 

^ = Si X 3-61 = 5.02% CaO - MgO, 

21.55% CaO present, 
48.60% CaO total. 

Calculating the AI2O3 into equivalent Si02, as it is prob- 
ably present as an aluminate replacing silica:^ 

* On the supposition that the mineral contains the alumina 
combined as an aluminate and the silica as a silicate, then the 
aluminate can be represented as 3 M"0 • AI2O3 and the silicate as 
M''0 • Si02; whence it follows that as AI2O3 requires three M"0 
and the SiOa one, hence AI2O3 = 3 Si02. 



ATOMIC WEIGHTS AND FORMULAS 161 

m = m?X°-«^= 1.50%SiO.^AlA, 

45.95 % SiOa present, 
47.45% Si02 total. 

^ ^ 48.60 „ „_ 0.866 ... . , i 

= ggQ9 = 0.866; ^-^ = 1.1 = 1 approximately. 

The mineral is of the type CaO • SiOa or CaSiOa or R° SiOa 
in which R" is chiefly Ca partially replaced by Mn, Zn, 
Fe° Mg, with a small amount of Si02 replaced by AI2O3, 
and is (Ca, Mn, Zn, Fe, Mg)Si03: (pyroxene). ^ This 
latter formula gives no indication as to the amounts of 
the metals present and does not show the alumina re- 
placement. 

1 The sum of the percentages of Si02 and CaO total 96.05%. 
The mineral as analyzed shows 0.31% loss on ignition. Deducting 
this leaves the analysis footing up to 101.56 — 0.31 = 101.25%. 
Calculating the percentage on this basis gives (see p. 176): 

47.45 



96.05 
48.60 
96.05 



X 101.25 = 50.02% Si02, 
X 101.25 = 51.23% CaO, 











101.25% total. 


Then 














Si02 


50.02 
60.3 


0.830; 


0.830 
0.830 


1, 




CaO 


51.23 
56.19 


0.915; 


0.915 
0.830 


1.1 = 1 approximately, 


and the ratio 


is as given before. 







162 CHEMICAL CALCULATIONS 

PROBLEMS 

277. The atom of oxygen is 1.1420 times as heavy as the 
atom of nitrogen, (a) Taking the atomic weight of oxygen to 
be 16, what is the atomic weight of nitrogen? (6) If oxygen 
were 100, what is the atomic weight of nitrogen? 

Ans. (a) 14.01. 
(6) 87.56. 

278. When oxygen is taken as 16, the atomic weight of iodine 
is 126.92 and hydrogen is 1.008. What is the atomic weight of 
iodine in a system taking hydrogen as unity? Ans. 125.91. 

279. Erdmann and Marchand heated mercuric oxide: 

2HgO = 2Hg + 02. 

From 118.3938 g. of mercuric oxide, 109.6308 g. of mercury were 
obtained. Calculate the atomic weight of mercury. 

Ans. 200.17. 

280. The atomic weight of lead was determined by Stas by 
converting lead into the nitrate: 100.00 g. of lead yielding 
159.9703 g. of the nitrate. What is the atomic weight of lead 
from these data? Ans. 206.80. 

281. Berzelius found that 2.265 g. of ferric oxide (Fe203) are 
formed from 1.586 g. of iron. Determine the atomic weight of 
iron from these data. Ans. 56.059. 

282. Barium sulphate contains 58.85% barium, 13.73% 
sulphur and 27.41% oxygen, (a) What is the atomic weight 
of barium when oxygen is 16 and sulphur is 32.07? (6) What 
is the atomic weight of sulphur when barium is 137.41 and 
oxygen 16? Ans. (a) 137.41. 

(b) 32.07. . 

283. The analysis of chromous chloride (CrCl2) according to 
Pelegot is 57.50% chlorine and 42.50% chromium. What is 
the atomic weight of chromium? Ans. 52.419. 

284. Lead sulphate on analysis yields 68.31% lead, 10.58% 
sulphur and 21.11% oxygen. What is the atomic weight of 
lead? Ans. 207.1. 



ATOMIC WEIGHTS AND FORMULAS 163 

285. According to Berzelius, litharge (PbO) contains 7.1724% 
oxygen. What is the atomic weight of lead? Ans. 207.08. 

286. Calculate the atomic weight of silver from the follow- 
ing figures determined by Stas. 53.1958 g. of silver form 
92.6042 g. of silver bromide. Ans. 107.88. 

287. Dumas passed hydrogen gas over heated copper oxide. 
The copper oxide lost 59.789 g.; the water formed weighed 
67.282 g. Calculate the atomic weight of hydrogen. 

Ans. 1.0026. 

288. Working with the reaction, 

2Ag + Cl2 = 2AgCl, 

Stas found that 91.4620 g. of silver produced 121.4993 g. of 
silver chloride. Calculate the atomic weight of chlorine. 

Ans. 35.428. 

289. According to the reactions, 

2 AuCls + 3 SO2 + 6 H2O = 2 Au + 6 HCl + 3 H2SO4 
3 H2SO4 + 3 BaCl2 = 3 BaSO^ + 6 HCl, 

Levol found that 1000 parts of gold in the form of the chloride 
produce an amount of sulphuric acid that will precipitate 1728 
parts of barium sulphate. Calculate the atomic weight of gold. 

Ans. 202.64. 

290. Chromic oxide (Cr203) contains 68.42% chromium and 
31.58% oxygen. What is the atomic weight of chromium? 

Ans. 52.00. 

291. Erdmann and Marchand determined the atomic weight 
of mercury by heating the sulphide with copper, 

HgS + Cu = CuS + Hg, 

and found 177.1664 g. of mercuric sulphide gave 152.745 g. of 
mercury. Calculate the atomic weight of mercury. 

Ans. 200.59. 

292. The percentage composition of platinum chloride is 

Per cent 

Platinum 57.92 

Chlorine 42.08 

100.00 



164 CHEMICAL CALCULATIONS 

Taking 35.46 as the atomic weight of chlorine and 0.0323 as 
the specific heat of platinum, what is the atomic weight and 
the valence of platinum? Ans. At. wt. = 195.20. 

Valence = 4. 

293. The percentage composition of ferrous oxide is 

Per cent 

Iron 77.73 

Oxygen 22.27 

100.00 

The specific heat of iron is 0.1162. What is the atomic weight 
and the valence of iron in this compound? 

Ans. At. wt. = 55.84. 
Valence = 2. 

294. Antimony forms compounds with oxygen of the follow- 
ing compositions: 

(a) 
Per cent 

Antimony 83 . 35 

Oxygen 16.65 

100.00 100.00 100.00 

The specific heat of antimony is 0.0495. What is the atomic 
weight of antimony and its valence in the compounds (a), (6) 
and (c)? Ans. At. wt. = 120.2. 

Valence =(a) 3. 

(6) 4. 

(c) 5. 

|295. Siewert found that 36.865 parts of chromic chloride 
gave 100 parts of silver chloride: 

CrCls + 3 AgNOg = Cr(N03)3 + 3 AgCl. 

Calculate the atomic weight of chromium from these data. 

Ans. 52.15. 

296. According to Seubert, when platinum tetrachloride 
(PtCli) is heated in hydrogen, hydrochloric acid is formed. 
This he precipitated with silver nitrate and weighed as the 



(&) 


(c) 


Per cent 


Per cent 


78.98 


75.03 


21.02 


24.97 



ATOMIC WEIGHTS AND FORMULAS 165 

chloride of silver. 17.4139 g. of silver chloride were found to 
be equivalent to 5.9242 g. of platinum. The reactions are 

PtCl4 + 2H2 = Pt + 4HCl 

HCl + AgNOa = HNO3 + AgCl. 

Calculate the atomic weight of platinum. Ans. 195.05. 

297. Berzelius determined the atomic weight of platinum by- 
converting the platinum into platinic chloride (PtCU) and 
finding the amount of potassium platinic chloride (K2PtCl6) 
formed, 

PtC]4 + 2KCl = K2PtCl6. 

24.735 g. of potassium platinic chloride were formed from 
10.000 g. of platinum. Calculate the atomic weight of platinum. 

Ans. 197.46. 

298. Berlin found that 100.000 g. of lead nitrate gave 97.576 g. 
of lead chromate, 

Pb(N03)2 + K2Cr04 = PbCr04 + 2 KNO3. 

Calculate the atomic weight of chromium. Ans. 51.99. 

299. Calculate the atomic weight of aluminum from the 
figures given by Mallet: 8.2144 g. of ammonium alum, 
Al2(S04)3-(NH4)2S04-24H20 gave 0.9258 g. of aluminum 
oxide (AI2O3). Ans. 27.113. 

300. Mallet found that 6.9617 g. of aluminum bromide re- 
quired 8.4429 g. of dissolved silver for precipitation : 

AlBrs + 3 AgNOs = Al (N03)3 + 3 AgBr. 

Calculate the atomic weight of aluminum. Ans. 27.093. 

301. Stas found that when he added 7.25682 g. of potassium 
chloride to 10.51995 g. of silver dissolved in nitric acid that 
0.01940 g. of silver remained in solution. The reaction being 

KCl + AgNOs = AgCl + KNO3, 
calculate the atomic weight of potassium. Ans. 39.09. 

302. Calculate the atomic weight of carbon in each case from 
the following figures of Ptendtenbacher and Liebig: (a) 28.803 g. 



166 CHEMICAL CALCULATIONS 

silver acetate (CHgCOOAg) yielded 18.612 g. silver. (6) 
16.220 g. silver tartrate (C4H406Ag,) yielded 9.6175 g. silver, 
(c) 25.898 g. silver malate (C4H406Ag2) yielded 16.059 g. silver. 

Ans. (a) 12.024. 

(6) 12.02. 

(c) 12.041. 

303. Marignac made the following determinations for the cal- 
cuhition of the atomic weight of nitrogen: (a) From 314.894 g. 
of silver nitrate he obtained 200.000 g. of silver. (6) 14.1100 g. 
of silver nitrate required 6.1910 g. of potassium chloride for 
complete precipitation, (c) 10.339 g. of silver dissolved in 
nitric acid required 5.1200 g. of ammonium chloride for com- 
plete precipitation. Determine the atomic weight of nitrogen 
in each case and {d) the mean atomic weight of nitrogen from 
these determinations. Ans. (a) 13.974. 

(6) 14.055. 
(c) 13.932. 
id) 13.987. 

304. 3.000 g. of iodine dissolved in 90 g. of ethyl ether raised 
the boiling point 0.283° C. What is its molecular weight? 

Ans. 254. 

305. If 1.32 g. of formic acid dissolved in 100 g. of water 
lowers the freezing point 0.53° C, what is its molecular weight? 

Ans. 46.1. 

306. 10.000 g. of naphthalene dissolved in 100 g. of benzol 
lowers the freezing point 3.90° C. What is the molecular 
weight? Ans. 128. 

307. 1.67 g. of benzoic acid dissolved in 70 g. of ether raised 
the boiling point 0.422° C. Calculate the molecular weight. 

Ans. 122. 

308. 0.200 g. of alcohol dissolved in 100 g. of benzol lowered 
the freezing point 0.210° C. What is the molecular weight of 
the alcohol? Ans. 46.3. 



ATOMIC WEIGHTS AND FORMULAS 167 

309. 1.000 g. of naphthalene dissolved in 50 g. of chloroform 
raised the boiling point 0.56° C. What is its molecular weight? 

Ans. 128. 

310. Calculate the molecular weight of nitrosodiphenyl amine 
if 3.58 g. of the substance dissolved in 150 g. of benzol lowers 
the freezing point 0.601° C. Ans. 198.5. 

311. Tartaric acid has two acidic hydrogen ions. When 
both these are replaced by silver the silver salt contains 59.31% 
silver. What is the molecular weight of tartaric acid assuming 
it contains no water of crystallization? Ans. 150.046. 

312. Silver acetate has one replaceable hydrogen ion. 1.0000 
g. of silver produces 1.5475 g. of silver acetate. What is the 
molecular weight of acetic acid? Ans. 60.073. 

313. The silver salt of fumaric acid (dibasic) contains 65.43% 
silver. What is the molecular weight of fumaric acid? 

Ans. 116.016. 

314. 2.0000 g. of the hydrochloro platinic acid salt of dimethyl 
anihne yield 0.5967 g. of platinum. What is the molecular 
weight of dimethyl anihne? Ans. 122.13. 

315. 1.0000 g. of the hydrochloro platinic acid salt of pyri- 
dine yields 0.3436 g. of platinum on ignition. Calculate the 
molecular weight of pyridine. Ans. 79.08. 

316. What is the molecular weight of ethyl amine if 3.0000 g. 
of the hydrochloro platinic acid salt yield 1.1708 g. of platinum 
on ignition? Ans. 45.10. 

317. An analysis of arsenic pentoxide shows 65.2% arsenic 
and 34.8% oxygen. Its molecular weight is 230. What is the 
formula? Ans. AS2O5. 

318. Arsenic trichloride analyzes 41.4% arsenic and 58.6% 
chlorine. Its molecular weight is 181. What is the formula? 

Ans. AsCls. 

319. Carbon trichloride shows a molecular weight of 237. 
Analysis shows it to contain 89.9% chlorine and 10.1% carbon. 
What is its molecular formula? Ans. C2CIC. 



168 CHEMICAL CALCULATIONS 

320. Carbanil contains 

Per cent 

Carbon 70.6 

Hydrogen 4.2 

Nitrogen 11.8 

Oxygen 13.4 

The molecular weight is 119. What is the formula? 

Ans. C7H5NO. 

321. Phenyl cyanide has a molecular weight of 103. Its 
composition is 

Per cent 

Carbon 81.5 

Hydrogen 4.9 

Nitrogen 13.6 

What is the formula? Ans. C7H5N. 

322. Phenyl disulphide has a molecular weight of 218. 
Analysis shows 70.6% CgHs and 29.4% sulphur. What is 
the formula? Ans. (C6H5)2S2. 

323. What is the empiric formula of a compound analyzing: 

Per cent 

Phosphorus 3.99 

Bromine 82.30 

Chlorine 13.70 

Ans. PBrgClg. 

324. What is the empiric formula of a compound of the com- 
position: 

Per cent 

Molybdenum 46.01 

Oxygen 1 1 . 50 

Chlorine 42. 49 

Ans. M02O3CI5. 

325. If basic antimonyl sulphate has the composition 

Per cent 

SbO 40.5 

SO4 14.2 

Sb 35.5 

OH 10.0 

what is the formula? Ans. (SbO) 2 SO4 • Sb2 (0H)4. 



ATOMIC WEIGHTS AND FORMULAS 169 

326. Find the empiric formula of a compound of 

Per cent 

Nickel 26. 12 

Hydrogen 0.90 

Arsenic 44 . 51 

Oxygen 28. 48 

Ans. NisHe (As03)4. 

327. Find the empiric formula of a compound of • 

Per cent 

Chromic oxide (Cr203) 15.9 

Sulphur trioxide (SO3) 33.5 

Ammonium oxide (NH4)20 5.4 

Hydrogen oxide (H2O) 45 . 2 

Ans. Cr2 (SO^s • (NH4)2S04 • 24 H2O. 

328. What is the formula of the following; lead being most 
accurately determined? 

Per cent 

Lead 83.61 

Sulphur 9.52 

Chlorine 7.36 

Ans. Pb4S3Cl2. 

329. Platinum is most accurately determined in the follow- 
ing. What is the formula? 

Per cent 

Lithium 2.91 

Platinum 36. 84 

Chlorine 40. 10 

Water 20. 60 

Ans. LisPtCle . 6 H2O. 

330. Calculate the simplest formula, considering arsenic oxide 
as the most accurately determined, of 

Per cent 

Arsenic oxide (AS2O5) 37. 91 

Calcium oxide (CaO) 18.28 

Ammonium oxide (NH4)20 8 . 70 

Water of crystallization 35 . 23 

Ans. CaNH4As04 • 6 H2O. 



170 CHEMICAL CALCULATIONS 

331. Calculate the empirical formula of the following sub- 
stance, considering calcium as being most accurately deter- 
mined. 

Per cent 

Calcium 39 . 01 

Phosphorus 19 . 85 

Oxygen (by diff.) 41 . 14 

Ans. Ca3P208. 

332. Find the formula of a substance analyzing 

Per cent 

Potassium 11 . 47 

Antimonyl (SbO) 40. 80 

Carbon 14 . 65 

Hydrogen 1.19 

Oxygen 29.08 

Water of crystallization 2 . 60 

considering the antimonyl most accurately determined. 

Ans. K2 (SbO) 2 CsHsOia • H2O. 

333. Given the following as the composition of Hypersthene, 
what is its formula? 

Per cent 

Si02 54.20 

FeO 21.70 

MgO 24.10 

100.00 

Ans. (Mg, Fe) SiOs. 

334. A sample of Smaltite examined by Hofmann gave the 
following analysis (disregard the Bi and S). What is its formula? 

Per cent 

As 70.37 

Co 13.95 - 

Ni 1.79 

Fe.. 11.71 

Cu 1.39 

S 0.66 

Bi 0.01 

99.88 

Ans. (Co,Ni,Fe,Cu) As2. 



ATOMIC WEIGHTS AND FORMULAS 171 

336. A sample of Wad analyzed: 

Per cent 

Mn02 59 . 94 

MnO 6.58 

ZnO 21 . 70 

FezOs 0.25 

H2O 11.58 

100.05 

Disregarding the iron, what is the formula? 

Ans. 2 Mn02 • (Zn,Mn) • 2 H2O. 

336. What is the formula of a sample of Diallage analyzed by 
Von Rath, and found to be (disregarding the water) : 

Per cent 

SiOz 53.60 

AI2O3 1.99 

FeO 8.95 

MnO '. 0.28 

MgO 13.08 

CaO 21.06 

H2O 0.86 

99.82 
Ans. (Ca,Fe,Al,Mn,Mg) SiOs. 

337. Show that the mineral analyzed below by Striiber 
(Gastaldite) approximates the formula R3Al4Si9027. 

Per cent 

SiOz 58.55 

AI2O3 21.40 

FeO 9.04 

MgO 3.92 

CaO 2.03 

Na20 4.77 

99.71 

338. Deduce a formula for natrolite which was analyzed by 
Harrington with the following results : 



172 CHEMICAL CALCULATIONS 

Per cent 

SiOz 47.09 

AI2O3 26.99 

NazO 16.46 

K2O 0.01 

H2O 9.80 

100.35 
Ans. Na20 • AI2O3 • 3 Si02 • 2 H2O. 

339. What is the formula of Caberite which, according to 
Sachs, analyzes as follows: 

Per cent 

AsaOfi 40.45 

NiO 26.97 

FeO 1.10 

MgO 6.16 

H2O 25.26 

99.94 

Ans. (Ni,Fe,Mg) ASO4 • 8 H2O. 

340. Wolframite, according to Finlayson, gave on analysis the 
following, what is its formula? 

Per cent 

. WO3 76.24 

FeO 16.39 

MnO 6.05 

CaO 1.05 

MgO 0.11 

99.84 

Ans. (Fe,Mn,Ca,Mg)W04. 



CHAPTER VIII 
GRAVIMETRIC ANALYSIS 

In a previous section, the amount of an element or 
radical obtainable from a given weight of a substance 
entering into a reaction was considered. It was postu- 
lated that the substances were pure, and only on this 
assumption are such calculations possible. 

Direct Gravimetric Analysis. — Gravimetric analysis 
consists of weighing an element or compound of known 
composition and from this weight calculating the amount 
of a constituent present in the original substance. As 
gravimetric analyses are reported as percentages of 
constituents determined, it is necessary to know the 
weight of substance taken for analysis. Gravimetric 
analysis presupposes a knowledge of the elements or 
radicals present in the substance to be analyzed; in other 
words, it is preceded by a qualitative examination when 
the component elements or radicals are unknown. 

As an example: A salt known to contain sodium and 
chlorine is to be examined as to its purity. One gram of 
the substance is weighed out and after proper precau- 
tions the chlorine is precipitated as silver chloride accord- 
ing to the reaction 

NaCl + AgNOa = AgCl + NaNOs, 

the silver chloride in pure state being found to weigh 
2.4382 g. The weight of chlorine present in the silver 
chloride is 

173 



174 CHEMICAL CALCULATIONS 

IgCi = ill X2-«82 = 0.6032 g.Cl, 

and the percentage of chlorine is, one gram of substance 
being taken, 

0.6032 

1.0000 

or in one equation 



X 100 = 60.32% CI, 



""' '^-^^xf^X 100 ^60.32% CI. 



( 



AgCl 143.34 'M. 0000 

The sodium of the sodium chloride is converted into pure 
sodium sulphate and weighed. The weight is found 
to be 1.2083 g. Then, as there are two atoms of sodium 
in sodium sulphate, 

2Na-»Na2S04, 

the per cent of sodium present is, 

2 Na 46.00 ,, 1 .2083 ,, ,^^ ^^ , ^^ , ^ 
X 77^7^ X 100 = 39.12% Na, 



Na2S04 142.07 '^ 1.0000 

and the percentage accounted for is 60.32% + 39.12% = 
99.44%. From these figures the compound is found to be 
NaCl. The impurities are present to the extent of 100.00% 
— 99.44% = 0.56%. A complete quantitative analysis 
would require the nature and the amount of these impu- 
rities. If it had been known in the beginning that the 
substance was sodium chloride, the amount of which 
in the sample is the end desired, the same could have 

1 An expression such as this is intended to mean that the fraction 
immediately following the equality sign has as its numerator and 
denominator the molecular weights of the symbols of the fraction. 
The equality sign cannot be interpreted as the equality sign of 
mathematics, but is meant only in this limited sense. The value 
of the fraction is the factor to use when AgCl is weighed, found or 
given and CI is sought or required. See Chem. Ann., pp. 10-36. 



GRAVIMETRIC ANALYSIS 175 

been calculated from the amount of silver chloride ob- 
tained, 

NaCl _ ^8^ 2,438_2 _ 

AgCl ~ 143.34 ^ 1.0000 ^ ^"" " ^^'^^^^ ^^^^• 

In the same way the percentage of a radical present may 
be calculated. Suppose one gram of a substance known 
to be a sulphate yields 1.2318 g. of barium sulphate and 
that it is desired to know the per cent of sulphuric anhy- 
dride present. This is 

^""^ ^^-^^ x!-i^X 100 = 42.25% SO3. 



BaS04 233.44 '^ 1.0000 

When a reaction produces a gas, the volume may be 
measured and reduced to standard conditions and the 
weight of the same calculated. Suppose 0.5000 g. of 
calcite is treated with sulphuric acid, 

CaCOa + H2SO4 = CaS04 + CO2 + H2O, 

and that the carbon dioxide measured over water at 15° C. 
and 762 mm. is 118.2 cc. To determine the per cent of 
calcium carbonate, assuming that there are no other car- 
bonates present: The volume of dry gas at standard (the 
tension of aqueous vapor at 15° C. being 12.73 mm.) is 

II X ^^^ ^60^'^^ X 0.1182 = 0.11046 L. CO., 

the weight of which is 

"■"S^ X 44.00 = 0.2170 g. CO2. 

This indicates the presence of 

CaCOa _ m07 0.2170 _ 

"COT ~ 44.00 ^ 05000 ^ ^"" ~ ^**-^"^° ^''^^" 



176 CHEMICAL CALCULATIONS 

or better, proceeding to the final result by one expression, 

273 762 - 12.73 0.1182 100.07 

288^ 760 ^ -^^^X^j^Xl00-9S.70%L^LO^. 

Elimination of a Constituent from the Analysis. — 

Substances may contain water, water of crystallization, 
water mechanically held in the sample, oil and other 
impurities. When such substances are analyzed the 
amount of such material is determined; but when report- 
ing, the analysis is often desired on the dry basis, or as 
if no such substances were present. In this case the con- 
stituents are reported with the same relative accuracy 
as was attained in the actual examination, excluding the 
substance it is desired to eliminate. A compound was 
analyzed and found to contain: 

Per cent 

Moisture 1 . 47 

Sodium oxide 57 . 42 

Carbonic anhydride 40.65 

99.54 

The analysis reported on the dry basis is 

99M-\a7 ^ ^^-^^2 = 58.28% Na.0, 

99.54% 

Factor Weights. — In the calculation of percentage 
compositions, a ratio is usually employed, such ratios 
being known as factors. If, then, a weight of substance 
is taken for analysis numerically equal to the factor which 
must be employed, the percentage composition is ascer- 
tained directly and is numerically equal to the mass of 
the substance separated and weighed; due cognizance 



GRAVIMETRIC ANALYSIS 177 

being taken of the position of the decimal point. The 
factor for the determination of sulphur equivalent to 

barium sulphate is ^ft^^t^ = f^f^tttj = 0.13738. Then 
^ BaS04 233.44 

if X units of mass of barium sulphate are produced from 
a quantity of material numerically equal to the factor 
0.13738, the percentage of sulphur is 

0.13738 X q3^ X 100 = 100 Z. 

The per cent of sulphur in the sample is given by the 
weight of the barium sulphate produced, the decimal 
point in this case being moved two points to the right. 
Weights numerically equal to the factor or to some simple 
multiple or submultiple of the factor which is to be em- 
ployed are called factor weights. 

Indirect Analysis. — Of indirect analysis there are so 
many different cases that all will not be treated individu- 
ally. A few types will be considered. 

Case I. — In which an equivalent amount of some 
other element or radical is determined. In the reac- 
tion 

AS2O3 + 2 H2O + 2 12 = 4 HI + AS2O5, 

the amounts of iodine necessary for the oxidation of arse- 
nious oxide to arsenic oxide being known, the amount of 
arsenious oxide or arsenic present is readily calculated. 
0.1014 g. of a sample containing arsenious oxide required 
0.2582 g. of iodine to oxidize to arsenic oxide. The per- 
centage of arsenious oxide is 

MO3 _ m92 02582 _ 

TT " 507.68 ^ 0.1014 ^ ^"" - ^^-^^ /'^ ^^^" 



178 CHEMICAL CALCULATIONS 

or if the percentage of the metal arsenic had been required, 
M^ 149^ 02582 
4 1 507.68 ^ 0.1014 ^ ^^^ ~ ^^'^^ /^ ^^' 

Case II. — Simple cases in which mixtures are weighed 
and an element or radical in it subsequently determined. 
Given the following: 

Amount of substance taken 5 . 0000 g. 

Which yields 1.0000 g. AI2O3 + FeaOa 

Which in turn yields 0. 1795 g. Fe 

to calculate the percentages of iron and aluminum in the 
sample. The percentage of iron is 

§:SSx 100 = 3.59% Fe. 
The amount of Fe203 equivalent to 0.1795 g. of Fe is 
YFq ^ 11168 ^'^^^^ = 0.2566 g. FesOs, 

then the amount of AI2O3 which was present in 1.0000 g. 
of the mixed oxides must have been 1.000 — 0.2566 = 
0.7434 g. AI2O3. The per cent of aluminum is 

2AL^ii2 07434 

AI2O3 102.2 ^ 5.0000 ^ ^^" " ^-^^ /^ ^^• 

Case III. — Simple cases in which mixtures are weighed 
and the mixture subsequently converted into a single 
compound. Given : 

Weight of substance taken 1 . 5000 g. 

Weight of mixture of AgBr + AgCl obtained . . 1 . 1060 g. 
Weight of AgCl produced from the mixture ^ . . . . 9563 g. 

to calculate the percentages of bromine and chlorine 
present in the original sample. 

1 The whole mixture AgCl + AgBr is converted into AgCl: 

X AgCl + y AgBr + | CI2 = (a: + y) AgCl + | Br2. 



GRAVIMETRIC ANALYSIS 179 

It is evident that an atom of chlorine has replaced an 
atom of bromine causing a loss of weight of 1.1060 — 
0.9563 = 0.1497 g. The loss in molecular weight corre- 
sponding to this is AgBr - AgCl = 187.80 - 143.34 = 
44.46, or what amounts to the same, the difference of the 
atomic weights of bromine and chlorine: 79.92 — 35.46 = 
44.46. The amount of bromine present which would 
cause this loss of weight is 
"Rr 1 7Q Q2 

BF^a = 4T46 >^ "-^^^^ = '^•2*'*^^ ^- ^'' 
and the per cent of bromine in the substance taken is 

^X 100 = 17.94% Br. 

The amount of silver bromide equivalent to the bromine, 
i.e., the amount that must have been present in the mix- 
ture, is 

^^ = ffx 0.2691= 0.6323 g. AgBr, 

or what amounts to the same, 

bSi = Sfx 0-^497 = 0.6323 g. AgBr. 

The weight of silver chloride present in the mixture is 
1.1060 - 0.6323 = 0.4737 g. The per cent of chlorine in 
the substance taken is 



AgCl 143.34 "" 1.5000 

Br 
^ The fraction -^ pY is meant to show that the fraction just 

after the equality sign is made up of numbers equal to the molec- 
ular weights of the elements symboHzed and that the operation has 
been carried out. In this case, 

Br ^ 70.92 ^ 70.92 

Br - CI ~ 70.92- 36.46 ~ 44.64" 



180 CHEMICAL CALCULATIONS 

To make this calculation algebraically, let 

X = weight of silver chloride present in the mixture, 
y = weight of silver bromide present in the mixture, 
a = weight of mixed silver salts, 
b = weight of mixed silver salts converted into silver 

chloride, 
/ = factor for the conversion of AgBr — > AgCl 
= AgCl_143^_ 

AgBr ~ 187.80 " "•^^^^^• 

Then 

x + y = a, 
x-\-fy = b, 

eliminating x by subtracting and solving for y, 

a — b 



1-/ 



y = 

substituting the value of /, 

_ a — b _ a — b _ , _, 

^ ~ 1 - 0.76325 ~ 0.23675 '"^'^^^^ ^^ ^^^ 
X = a — y. 
Applying to the problem given: 

a; + y = 1.1060 

x + 0.76325 2/ = 0.9563 



^ 1.1060 - 0.9563 
^ 1 - 0.76325 

= 0.6323 g. AgBr. 

Knowing that the reciprocal of 1 - 0.76235 is 4.2239, it 
is easier to proceed immediately to the result: 

y = 4.2239 (1.1060 - 0.9563) = 0.6323 g. AgBr, 
X = 1.1060 - 0.6323 = 0.4737 g. AgCl, 

and the percentages of chlorine and bromine in the sample 
taken are: 



GRAVIMETRIC ANALYSIS 181 

Br 79.92 ^O^^^^Q^^^^g^^^g^^ 



AgBr 187.80 1.5000 
CI 35.46^04737 ^^^^^^^^^^^^^ 



AgCl 143.34^^ 1.5000 

Instead of calculating the chlorine and bromine from the 

weights of the silver chloride and silver bromide, the 

amounts of chlorine and bromine can be determined 

Br . 79.92 
directly, as, for instance, the factor of . ^ being „ > 

the weight of bromine (z) is 
79.92 



z = 



X 4.2239 (a - h) 



187.80 
= 1.7975 {a - b). 

Multiplying factors are given in chemists' handbooks, 
the above being a method by which they are calculated.^ 
To calculate the factor for the chlorine: 

x+fy = b, 
x+ y = a, 

b-fa. 

substituting the values of /, 

a;=^^~^-'^^^^^^:=.4.2239(6-0.76325a)-4.22396-3.2239a. 
0.23675 ^ 

The factor to determine chlorine in silver chloride being 

then, if w is the weight of chlorine, 

w = 0.24738 (4.2239 b - 3.2239 a) = 1.0449 b - 0.79754 a. 

1 This same factor could have been obtained in the method pre- 

Br 79.92 

ceding the algebraic method: ^ _ ^ = |^^ = 1.7975. 



182 CHEMICAL CALCULATIONS 

Case IV. — Complex cases in which a known weight 
of the mixture is subsequently converted into a single 
element or a radical. Given: 

Weight of KCl + NaCl in mixture 1.0000 g. 

Weight of CI in the mixture subsequently determined. . 0.5411 g. 

it is required to determine the weights of sodium chloride 
and potassium chloride present in the mixture. Several 
methods will be given. 

Method (a). — If the mixture of KCl + NaCl had been 
pure NaCl the amount of chlorine would be 

NSl = S:Sx^-0000 = 0-'^0S57g.Cl; 

if it had been pure KCl the amount of chlorine would be 

The difference between the chlorine content if the mixture 
had been pure sodium chloride or pure potassium chloride 
is 0.60657 - 0.47559 = 0.13098 g. The observed differ- 
ence due to the presence of potassium chloride in the mix- 
ture is 0.60657 - 0.5411 = 0.06547 g. The amount of 
potassium chloride in the mixture must be, therefore, 

?^5gg X 1.0000 = 0.4999 g. KCL 

Then the weight of sodium chloride present in the mix- 
ture is 1.0000 - 0.4999 = 0.5001 g. Knowing the amounts 
of sodium chloride and potassium chloride, also the weight 
taken, the percentages are calculated in the usual way. 

Method (b). — If the mixture, were pure potassium 
chloride, the weight of the mixture equivalent to 0.5411 g. 
of chlorine is 

KCl 74.56 



CI 35.46 



X 0.5411 = 1.1378 g. 



GRAVIMETRIC ANALYSIS 183 

1.1378 - 1.0000 = 0.1378 g., excess due to the presence 
of sodium chloride. The amount of sodium chloride cor- 
responding to this difference is 

NaCl ^ 58.46 _ 58.46 
KCl - NaCl 74.56 - 58.46 16.10 ^ 
= 0.5004 g. NaCl, 
1.0000 - 0.5004 = 0.4996 g. KCl. 
Method (c). — Let 
X = weight of KCl in mixture, 
y = weight of NaCl in mixture, 
a = weight of mixture, 
h = weight of CI in mixture, 
/ = factor to determine the amount of CI in KCl 

= 0.47559, 
q = factor to determine the amount of CI in NaCl 

= 0.60657. 
Then 

X -{- y = a, 
fx + qy = h. 

Substituting the values for the particular problem under 
consideration : 

x+ 2/= 1.0000 

0.47559 X +0.60657 y = 0.5411 

X = 0.4999 = g. KCl 

2/ = 1.0000 -0.4999 = 0.5001 =g. NaCl. 

Another problem will be cited but slightly different 
from the foregoing. A sample containing potassium 
iodide and potassium bromide is to be examined for per- 
centage composition. One gram of the sample yields 
1.5370 g. of mixed silver bromide and silver iodide. This 
mixture is now converted into silver chloride and weighs 
1.1191 g. Although this latter compound contains none 
of the elements sought, these can be readily calculated. 



184 CHEMICAL CALCULATIONS 

If the mixture weighed had been pure silver iodide, the 
amount of chloride would have been 

i = 231:80 X ^-5370 = 0.9383 g.AgCl. 

If the mixture had been pure silver bromide, the amount 
of silver chloride would have been 

^r=;-ftSxi-53^0= 1-1^31 g.AgCl. 

1.1191 - 0.9383 = 0.1808 g., excess of silver chloride be^ 
yond what it would be if it were pure silver iodide, the ex- 
cess being due to the presence of silver bromide. 1.1731 — 
0.9383 = 0.2348 g., the difference in weight between what 
the silver chloride would weigh if derived from pure 
silver iodide or pure silver bromide. The weight of 
silver bromide present amounts to 

1 SOS 

^i^X 1.5370== 1.1835 g.AgBr, 

and 

1.5370 - 1.1835 = 0.3535 g.Agl. 

The percentages of the potassium bromide and potassium 
iodide in the original sample are: 

KBr _ IIM? X ^^^ X 100 - 7500'7 KBr 
AiB? ~ 187.80 X 1.0000 ^ ^"" " ^^""'^ ^^'^' 

KI _ 166.02 0.3535 _ ^ 

Ail - 234.80 X 1.0000 ^ ^"" ~ ^^'^^^ ^^■ 

Algebraically, the factors being 

Agl -> AgCl = ^^ = 0.61047, 

AgBr -^ AgCl = ^' = 0.76325; 



then, if 



Solving, 



GRAVIMETRIC ANALYSIS 185 



X = weight of AgBr, 
y = weight of Agl, 

x-\- y = 1.5370, 

0.76325 X + 0.61047 y = 1.1191. 



X = 1.1835 g. AgBr, 
y = 1.5370 - 1.1835 = 0.3535 g. Agl. 
The percentages are calculated as given before. 

The calculation of a factor in this case is as follows : let 
X = grams AgBr present in the mixed salt, 
y = grams Agl present in the mixed salt, 
a = grams mixed salt, 

b = grams AgCl obtained from the mixed salt, 
/ = factor for the conversion : AgBr — » AgCl, 
_ AgCl _ 143.34 

k = factor for the conversion: Agl— > AgCl; 
= AgCl ^143^^ 

Agl 234.80 ^•^^^^^• 
Then 

x-{- y = a, 
fx-^ky = h, 

fa-h 

Substituting the values of / and k, 

y = 4.9959 a -6.5454 6. 

If it is desired to know the amount of iodine in the mixed 

salts, the factor of iodine equivalent to silver iodide being 

I 126 92 

1^^ = ^o/o^ = 0.54054, then if o is the weight of iodine 
Agl 234.80 ' ^ 

present, 

q = 0.54054 (4.9959 a - 6.5454 b) 

= 2.7005 a - 3.5381 b. 



186 CHEMICAL CALCULATIONS 

The above is a general method^ Having the value of y 
for any mixed salt of silver bromide and silver iodide, a 
factor such as q may be calculated for any desired element, 
radical or compound known to be present in the original 
sample. 

Case V. — In which a mixture containing two salts 
having an element or radical in common is converted into 
another mixture, also containing an element or radical 
in common. Given, one gram of a mixture of sodium 
chloride and potassium chloride; the mixture being con- 
verted by treatment with sulphuric acid into a mixture 
of sodium sulphate and potassium sulphate, the weight 
of this latter mixture being 1.18600 g. It is required to 
calculate the weights of sodium chloride and potassium 
chloride present in the original mixture. If the sub- 
stance had been pure sodium chloride, the weight of so- 
dium sulphate obtainable would be 

Na2S04 142.07 



2NaCl 116.92 



X 1.00000 = 1.21510 g.Na2S04. 



If it had been pure potassium chloride, the weight obtained 
from one gram would be 

SSl = S5 ^ ^-^^ = 1.16866 g. K,S04. 

The difference between these weights is 

1.21510 - 1.16866 = 0.04644 g. 

The difference between the weight found and the weight 
which would have been obtained if the substance had been 
pure sodium chloride is 

1.18600 - 1.16866 = 0.01734 g. 

^ For a tabulation of indirect factors, see Chem. Ann., pp. 37-38. 



GRAVIMETRIC ANALYSIS 187 

Therefore, the amount of sodium chloride present in tho 
mixture of chlorides is 

^'^',, X 1.00000 = 0.37338 g. NaCl. 
0.04644 * 

By the same line of reasoning, the potassium chloride is 
1.21510 - 1.18600 = 0.02910. 

^;5gO X 1.00000 = 0.62662 g. KCl, 

or, by difference, 

1.00000 - 0.37388 = 0.62662 g. KCl. 

To calculate algebraically, let x = weight of NaCl, 
then 1 — a; = weight of KCl. 

m = factor ^^^^' = 1.2151044, 
2 NaCl 

n = factor -p^p, = 1.168656. 

1.215104 a; + 1.168656 {1 - x) = 1.18600, 
1.215104 x+ 1.168656 -'1.168656 x = 1.18600, 
0.046448 a; - 0.01734, 

- = 0^W = 0-3^338 g. NaCl. 

1.00000 - 0.37338 = 0.62662 g. KCl. 
To calculate the factor, let 

X = weight of sodium chloride, 
y = weight of potassium chloride, 
a = weight of mixed sulphates, 
b = weight of mixed chlorides, 

m = factor I^^J = 1.2151044, 
n = factor l^g^ = 1.168656. 



188 



CHEMICAL CALCULATIONS 



y 



1.215104 a; + 1.168656 {b - x) = a, 
1.215104 X + 1.168656 6 - 1.168656 x = a, 
0.046448 X = a- 1.168656 6, 

= ^-^Jf^^^^f ^ = 21.5293 {a - 1.168656 6) 
0.046448 ^ ^ 

= 21.5293 a- 25.1604 6. 

1.168656 2/ + 1.215104 (b - y) = a, 
1.168656?/ + 1.215104 6 - 1.215104?/ = a, 

-0.046448?/ - -1.215104 6 + a, 

1.215104 6 - a ^i 5293 (1.215104 6 - a) 



0.046448 
= 26.1604 6 -21.5293 a. 
To calculate this factor in another manner, let 
X = weight of sodium chloride, 
y = weight of potassium chloride, 
a = weight of mixed sulphates, 
6 = weight of mixed chlorides, 

m = factor ^^S^' = 1.2151044, 



n = factor 



2NaCl 

K2SO4 



1.168656. 



2KC1 
Factor for sodium chloride: Factor for potassium chloride: 



x + 


y 


= h, 


mx + 


ny 


= a, 


nx + 


ny 


= nh, 


mx H- 


ny 


= a, 


(m — 


n) X = a — 


a — 
x= 

m - 


nh 

~n' 


1 


- a - 


n 



m — n 



m 



x+ y = h, 
mx + ny = a, 



mx + my 
mx + ny 



mh, 
a, 



(m ■ 

y 
y 



n) y = mh — a, 
mh — a 
m — n ' 

m 
m — n 



6- 



m — n 



a. 



GRAVIMETRIC ANALYSIS 189 



Substituting the values of m and n, 



1 1.16866 , 



0.046448 0.046448 
X = 21.5293 a- 25.1604 6 



^ 1.215104 1 

^ 0.046448^ 0.046448 ' 

y = 26.1604 6 -21.5293 a. 



An inspection of these factors shows that an error made 
in weighing the mixed chlorides or mixed sulphates is 
multiplied more than twenty times in calculating the 
results. This should serve as a warning against using an 
indirect method of analysis when a direct method may be 
made to serve. Such methods of indirect analysis, beauti- 
ful as they may appear on paper, are to be avoided, ex- 
cept, possibly, in those cases in which the multiplying 
factor is small. The errors are less when approximately 
equal weights of the two substances are present. They 
may become very large when one of the substances is 
present to a much larger extent than the other. 

A table of calculations is given on page 190 showing the 
errors introduced by only one milligram error in weighing 
the mixed chlorides or sulphates. From an inspection of 
this table, it is clear that this particular method, at least, 
is little short of useless, except in exceptional cases and 
when the weights of the two chlorides are about equal. 



190 



CHEMICAL CALCULATIONS 









^ o 




o o 




i-H Oi 


-g 






• (M_ CO 


<M lO 




■ oi CO 


«, s 


y 




lO Ttl 


d ,-i 




(M C^ 


1 1 


W 




• + 1 


■ + 1 




• + 1 




'xh o 


O Ci 


o o 1 


? 




O CO 




• 00 CO 




(N to 

lO r-H 


2 


;zi 












(M (M 


<o 




• 1 + 


• 1 + 


• 1 + 






(N lO 


(M to 


oq to 






CO T-H 


• CO i-t 


CO l-H 


M ■ 


(N <N 


'■ (N d 


• (rq d 


P 


• + 1 


• + 1 


• + 1 




(M lO 


(M to 


cq to 


£ ** 


o 


lO .-1 


to l-H 


to r-( 


(1| 


1 


• c^ cq 


• cq d 


ci (m' 




■ 1 + 


• 1 + 


• 1 + 






O (M O 


O CQ to 
O CO CX) 


O (M to 


is 


p 


o CO 00 


O CO 00 


§,0, 


d ci t^ 


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GRAVIMETRIC ANALYSIS 191 

Mixture of Sulphuric Acid and Sulphuric Anhydride 
(Oleum). — Another case of mixtures is the acids and 
their anhydrides. The commonest of these is "oleum'' 
or fuming sulphuric acid, being a mixture of H2SO4 and 
varying amounts of SOs.^ In commerce an oleum is 
spoken of as being 30% oleum, for example, when it con- 
tains 30% by weight of free sulphur trioxide, the remainder 
being sulphuric acid, the per cent of free sulphur trioxide 
serving to designate the per cent oleum. 

Suppose it is required to know the total per cent of sul- 
phur trioxide in a sample of oleum. Let 

X = per cent oleum, i.e., the per cent uncombined SO3, 

100 — X = per cent sulphuric acid, 

then as the factor ^^ ^^ = ^ ' = 0.8163, if y is the 
xi2fe^4 Uo.uy 

per cent total SO3, 

y = 0.8163 (100 - x) + x. 

Or being given the per cent total SO3, to find the per cent 
uncombined SO3, i.e., the per cent oleum, from the equa- 
tion above is obtained 

x = 5.4436 (y - 81. 63). 2 

Examples. — (a) What is the percentage total SO3 in 
30% oleum? (5) A sample of oleum analyzes 85.31% 
SO3. What per cent oleum is it? 

(a) y = 0.8163 (100 - 30) + 30 

= 87.14% SO3. 

(b) X = 5.4436 (85.31 - 81.63) 

= 20.03% oleum. 

1 As SO3 has a great affinity for water, oleum can contain no H2O 
as water. 

2 For a table of percentages of oleum corresponding to percent- 
ages total SO3, see Chem. Ann., pp. 397-398. This table may be 
calculated by the method given. 



192 CHEMICAL CALCULATIONS 

PROBLEMS 

341. 0.2017 g. of a substance containing potassium yields 
0.7092 g. of potassium platinic chloride. What is the per cent 
of potassium? Ans. 56.56%. 

342. A substance known to be largely NaHS04 • H2O is ex- 
amined as to its purity; 0.3062 g. of it produces 0.5093 g. of 
barium sulphate. What is the per cent of acid sodium sul- 
phate? Ans. 98.40%. 

343. Calcium fluoride was analyzed by precipitating the 
fluorine as barium silicofluoride (BaSiFe) and the calcium was 
estimated as the oxide. 0.5042 g. of the fluoride gave 0.6002 g. 
of barium siHcofluoride and 0.3630 g. of calcium oxide. What 
is the per cent composition according to these figures? 

Ans. 48.52% F. 
51.45% Ca. 

344. A substance is suspected of being phosphorus pent- 
oxide (P2O5). 0.5043 g. of the substance dissolved in water 
gave 0.7835 g. of magnesium pyrophosphate (Mg2P207). Was 
the substance phosphorus pentoxide? Ans. Yes, 99.10%. 

345. A sample of a substance is known to be an oxide of 
arsenic. 0.2504 g. gave 0.3912 g. of magnesium pyroarsenate 
(Mg2As207). Is the substance arsenic oxide (AS2O5) or arsenious 
oxide (AsoOs)? Ans. 99.57% AS2O3. 

346. A sample containing barium, chlorine and water of 
crystallization was analyzed and the following figures obtained: 

Weight of sample taken 1 . 0000 g. 

Weight after heating (water driven off) . 8522 g. 

Weight of barium sulphate obtained 0. 9594 g. 

Weight of silver chloride obtained 1 . 1735 g. 

(a) Calculate the percentage of each constituent; (6) the 
formula of the substance and (c) the per cent errors on the 
assumption that the sample was pure. 

Ans. (a) 14.78% H2O; 56.46% Ba; 29.03% CI. 
(6) BaCl2.2H20. 
(c) H2O, + 0.03%; CI, ± 0; Ba, + 0.23%. 



GRAVIMETRIC ANALYSIS 193 

347. Potassium chlorate was analyzed as follows: the oxy- 
gen determined by loss of weight on heating; one gram lost 
0.3900 g. The chlorine determined by precipitation as silver 
chloride; one gram producing 1.1594 g. silver chloride. The 
potassium was weighed as potassium sulphate; one gram giving 
0.7098 g. of potassium sulphate, (a) Find the per cent com- 
position from these figures and (6) the formula. 

Ans. (a) 39.00% 0. 
28.68% CI. 
31.85% K. 
(b) KCIO3. 

348. A sample is analyzed for sodium oxide (Na20) and 
carbonic anhydride (CO2). 1.3417 g. of sodium sulphate are 
obtained from one gram of substance. The same amount pro- 
duces 229 cc. of carbonic anhydride measured moist at 17° C. 
and 757 mm. (a) What per cent of these constituents are 
present and (6) what is the formula of the compound? (Ten. 
aq. vapor at 17° C. = 14.45 mm.) Ans. (a) 58.55% NasO. 

41.37% CO2. 
(b) NasCOs. 

349. One gram of a solution of hydrogen dioxide is boiled 
liberating oxygen 

2H202 = 2H20 + 02. 

25.7 cc. of oxygen are produced measured moist at 758 mm. and 
15° C. What is the per cent of hydrogen dioxide in the solution? 
(Ten. aq. vapor at 15° C. = 12.73 mm.) Ans. 3.69%. 

350. A coal analyzes as follows: 

Per cent 

Volatile combustible matter including sulphur. . . 18. 10 

Fixed carbon 74 . 52 

Sulphur 0. 60 

Ash 6.65 

Water 0.73 

100.60 
What are these percentages on the dry basis? 

Ans. 18.23%; 75.06%; 0.61%; 6.70%. 



194 CHEMICAL CALCULATIONS 

351. A coke analyzes: 

Per cent 

Volatile combustible matter 1 . 58 

Fixed carbon 88. 87 

Sulphur 1 . 18 

Ash 8.99 

Water 1.92 

102.54 
What is its composition on the dry basis? 

Ans. 1.61%; 90.56%; 1.20%; 9.16%,. 

352. The extracted pigment of a paint analyzes: 

Per cent 

White lead 30.42 

Zinc oxide 61 . 52 

Asbestine 7 . 23 

Unextracted oil 1.06 

100.23 
Calculate the percentages present if all the oil had been extracted. 

Ans. 30.74%; 62.18%; 7.31%. 

353. A sample of phosphate rock analyzes: 

Per cent- 
Calcium phosphate 96 . 02 

Ferric oxide 1.13 

Aluminum oxide . 60 

Insoluble 1 . 24 

Moisture 0.73 

99.90 
Calculate the percentages on the dry basis. 

Ans. 96.91%; 1.14%; 0.60%; 1.25%. 

354. On the dry basis, a sample of coal analyzes: 

Per cent 

Volatile combustible 21 . 06 

Fixed carbon 71 . 80 

Ash 7.14 

100.00 
If the moisture present in the coal was 2.49%, what is the 
analysis on the wet basis? Ans. 20.54% 

70.01% 

6.96% 

2.49% 

100.00% 



GRAVIMETRIC ANALYSIS 195 

355. A sample of coal was dried at 70° C, at which tempera- 
ture a loss of 6.03% water was obtained. The partially dried 
sample was analyzed and showed: 

Per cent 

Volatile material 8. 23 

Fixed carbon 65 . 90 

Ash 23.80 

Moisture 2.07 

100.00 
What are the percentages on the original sample? 

Ans. 7.73% 

61.93% 

22.36% 

7.78% 

100.00% 

356. The pigment of a certain paint has the composition: 

Per cent 

(a) White lead 33. 30 

(6) Zinc oxide 66.70 

100.00 
The vehicle of this paint consists of 

Per cent 

■ (c) Linseed oil 32. 40 

(d) Japan dryer 13 . 60 

(e) Asphaltum spirits 54 . 00 

100.00 

The pigment and vehicle are mixed to form the paint in the 
proportions : 

Per cent 

Pigment 63.00 

Vehicle 37.00 

100.00 
Calculate the percentage composition of this paint. 

Ans, (a) 20.98%. 

(b) 42.02%. 

(c) 11.99%. 

(d) 5.03%. 

(e) 19.98%. 



196 CHEMICAL CALCULATIONS 

357. A paint has the following composition: 

Per cent 

(a) White lead 40.00 

(6) Zinc oxide 23 . 00 

(c) Linseed oil 15 . 00 

(d) Japan dryer 5 . 00 

(e) Asphaltum spirits 17 . 00 

TooToo 

(a) What is the percentage composition of the pigment? 
(6) What is the percentage composition of the vehicle? 

Ans. (a) (a) 63.49%. 

(b) 36.51%. 

(6) (c) 40.55%. 

(d) 13.51%. 

(e) 45.95%. 

358. Phosphoric anhydride (P2O5) is to be weighed as silver 
phosphate (Ag3P04). What is the factor weight to take in order 
that the weight of silver phosphate obtained multiplied by 10 
shall give the per cent of phosphoric anhydride? 

Ans. 1.6968 g. 

359. The percentage of phosphoric anhydride (P2O5) is to be 
determined as magnesium pyrophosphate (Mg2P207). What 
weight in grams must be taken in order that the weight of 
the pyrophosphate multiplied by 100 shall give the percentage 
of phosphoric anhydride? Ans. 0.6379 g. 

360. Bicarbonate of soda (NaHCOs) is to be examined as to 
its purity. The sodium is converted into sodium sulphate and 
weighed as such. What weight of bicarbonate must be taken 
in order that the weight of sodium sulphate obtained multiplied 
by 100 shall give the per cent of bicarbonate? 

Ans. 1.1827 g. 

361. Manganese dioxide (Mn02) is to be determined as tri- 
manganese tetroxide (Mn304). What is the factor weight 
order that the weight of tetroxide in grams multiplied by 1( 
shall give the percentage of manganese dioxide? 

Ans. 11.3987 g. 



GRAVIMETRIC ANALYSIS 197 

362. Sodium ammonium phosphate (NaNH4HP04 • 4 H2O) is 
to be analyzed. The phosphorus is precipitated as magnesium 
pyrophosphate (Mg2P207). What is the factor weight in order 
that the weight of the pyrophosphate in centigrams shall show 
the percentage of sodium ammonium phosphate? 

Ans. 1.8781 g. 

363. What is the per cent of sulphur dioxide in a sample of 
sodium sulphite when 0.1063 g. of the same requires 0.1615 g. of 
iodine to oxidize it according to the reaction: 

Na^SOa + I2 + H2O = Na2S04 + 2 HI? 

Ans. 35.95%. 

364. One gram of a sample containing an iodide is treated 
with chlorine, 

2MI + Cl2 = 2MCH-l2, 
and the iodine liberated is treated with sodium thiosulphate, 

2 NasSsOs + I2 = 2 NaU- Na2S406, 

0.9213 g. of the thiosulphate being required. What is the 
per cent of iodine in the sample? Ans. 73.94%. 

365. 1.5000 g. of a sample give 0.6214 g. of mixed strontium 
and calcium sulphates. From this mixture 0.4120 g. of calcium 
carbonate are obtained. Assuming calcium chloride and stron- 
tium chloride to have been present as such in the original sample,, 
calculate the percentages of these compounds. 

Ans. 30.46% CaCl2. 
3.50% SrCl2. 

366. From the following calculate the percentages of iron and 
aluminum: 

Weight of substance taken .... 1 . 0043 g. 

Which yields 0.2716 g. AI2O3 + Fe203 

Which in turn yields 0.0726 g. Fe. 

Ans. 7.23% Fe. 
8.86% Al. 

367. One gram of sample gives 0.1485 g. of mixed calcium 
and strontium carbonates, from which mixture 0.0957 g. of 



198 CHEMICAL CALCULATIONS 

strontium sulphate are obtained. What are the percentages of 
calcium oxide and strontium oxide? Ans. 5.40% SrO. 

4.01% CaO. 

368. Calculate the percentages of sodium and potassium, 
given the following data: 

Weight of substance taken 2 . 0000 g. 

Weight of NaCl + KCl obtained 0. 6437 g. 

Weight of K2PtCl6 obtained from mixture 0.6514 g. 

Ans. 5.24% K. 
8.92% Na. 

369. A substance contains sodium and potassium. It gives 
36.42% of mixed sodium and potassium sulphates. From the 
mixed sulphates is obtained a weight of potassium platinic 
chloride (K2PtCl6) that amounts to 47.20% of the weight of 
sample taken. What are the percentages of sodium and potas- 
sium oxide? Ans. 9.15% K2O. 

8.51% NaaO. 

370. 1.5000 g. of a substance containing bromine and chlo- 
rine yield 1.0000 g. of mixed silver chloride and silver bromide. 
The mixture is treated with chlorine, converting the silver bro-j 
mide into silver chloride, the loss of weight due to this change) 
being 0.1367 g. What are the percentages of chlorine and] 
bromine? Ans. 16.38% Br. 

6.97% CI. 

371. What are the percentages of chlorine and iodine in a 
substance, 2.0000 g. of which yield 1.5230 g. of mixed silver chlo- 
ride and iodide, which mixture when converted into the chloride 
yields 1.1090 g. of silver chloride? Ans. 28.73% I. 

15.69% CI. 

372. A sample composed of potassium chlorate (KCIO3) and 
potassium bromate (KBrOa) is examined by the indirect method. 
One gram was taken from which 1.0474 g. of mixed silver chlo- 
ride and bromide were obtained. From this mixture 1.0138 g. 
of silver chloride were obtained. What are the percentages 



GRAVIMETRIC ANALYSIS 199 

of potassium chlorate and potassium bromate in the original 
sample? Ans. 12.62% KBrOs. 

77.42% KCIO3. 

373. If 

X = weight of AgCl present, 

y = weight of Agl present, 

a = weight of mixture of AgCl + Agl, 

b — weight of AgCl obtained from (a) : 

(a) Calculate the factor which multiplying (a — b) gives the 
weight of silver iodide, and (6) the factor that multiplying 
(a — b) gives the weight of iodine. Ans. (a) 2.5672 {a — b). 

(b) 1.3877 (a -6). 

374. Taking the factors of problem 373, (a) what are the 
amounts of silver chloride and silver iodide when the mixed 
salts weigh 1.8320 g. and the mixed salts converted into the 
chloride weigh 1.5232 g.? (b) What is the weight of iodine 
present? Ans. (a) 0.7928 g. Agl. 

1.0392 g. AgCl. 
(6) 0.4285 g. I. 

375. What is the factor to be used which multiplying the loss 
of weight caused by converting a mixture of silver chloride and 
silver bromide into silver chloride will give the amount of bro- 
mine present? . Ans. 1.7976. 

376. What is the factor to be used which multiplying the loss 
of weight caused by converting a mixture of silver chloride and 
silver iodide into silver chloride will give the weight of iodine? 

Ans. 1.3877. 

377. One gram of a mixture of silver iodide and silver bro- 
mide is converted into silver chloride, yielding 0.6487 g. of the 
same. What are the weights of iodine and bromine in the mix- 
ture? Ans. 0.1065 g. Br. 

0.4053 g. I. 

378. 1.0721 g. of a mixture of calcium carbonate and stron- 
tium carbonate yield 1.4217 g. of a mixture of strontium sul- 



200 CHEMICAL CALCULATIONS 

phate and calcium sulphate. How much calcium carbonate and 
strontium carbonate were contained in the mixture? 

Ans. 0.7574 g. CaCOa. 
0.3148 g. SrCOa. 

379. 0.5541 g. of a mixture of potassium iodide and bromide 
yield 0.3396 g. of potassium sulphate. How much iodine and 
bromine in the mixture? Ans. 0.2436 g. I. 

0.1581 g. Br. 

380. A mixture of potassium sulphate and sodium sulphate 
weighs 1.4304 g. From this mixture 2.1364 g. of barium sul- 
phate are obtained, {a) What are the weights of potassium 
sulphate and sodium sulphate in the mixture? (6) If 1.5000 g. 
of the substance were taken from which this mixture was ob- 
tained, what are percentages of sodium and potassium in the 
original sample? 

Ans. (a) 0.7257 g. Na2S04. 
0.7047 g. K2SO4. 
(b) 15.67% Na. 
21.08% K 

381. Calculate the factors that applied to the previous prob- 
lem (380) will give the weights of potassium and sodium. These 
are to be multiplying factors. 

Ans. K = 2.4278 a- 1.4775 6, 
Na = 1.0662 6- 1.4280 a. 

382. One gram of a mixture consisting of silver chloride and 
silver bromide is found to contain 0.6635 g. of silver, (a) 
How much bromine and (b) how much chlorine does it contain? 

Ans. (a) 0.2128 g. Br. 
(6) 0.1237 g. CI. 

383. Calculate factors which when multiplying the weight of- 
the mixed salt of problem 382 and the weight of silver obtained 
from the mixture will give the weights of bromine and chlorine 
present. Ans. CI = 1.3884 b - 0.79753 a. 

Br = 1.7975 a - 2.3883 b. 

384. A mixture of barium and strontium carbonates weighing 
1.2601 g. contains 0.3192 g. of carbon dioxide, (a) What are 



GRAVIMETRIC ANALYSIS 201 

the weights of barium carbonate and strontium carbonate in 
the mixture? (6) What are the weights of barium and stron- 
tium in the mixture? Ans. (a) 0.7504 g. BaCOs. 

0.5097 g. SrCOs. 
(6) 0.5223 g. Ba. 

0.3025 g.Sr. 

385. Calculate the factors that applied to a problem such as 
384 will give the weights of barium and strontium. 

Ans. Ba - 2.7611 a - 9.2661 b, 
Sr = 7.9025 6- 1.7617 a. 

386. A sample of oleum weighing 3.1402 g. contains 2.7350 g. 
of sulphur trioxide. What per cent oleum is it? 

Ans. 29.78% oleum. 

387. What is the per cent total sulphur trioxide in 25% oleum? 

Ans. 86.22%. 

388. What is the per cent total sulphur trioxide in 15% oleum? 

Ans. 84.38%. 

389. If acetic anhydride (CH3CO)2 will take up any water 
present to form acetic acid, what is the amount of uncombined 
acetic anhydride present in a sample of mixed acetic acid and 
acetic anhydride showing 87.72% anhydride present? 

Ans. 18.13%. 



CHAPTER IX 

VOLUMETRIC ANALYSIS 

"Volumetric analysis or quantitive chemical analysis 
by measure in the case of liquids and solids . . . depends 
upon the following conditions for its successful practice: 

1. A solution of the reagent, the chemical value of 
which is accurately known, called the 'standard solu- 
tion. ' 

2. A graduated vessel from which portions of it may be 
accurately delivered, called the 'burette.' 

3. The decomposition ^ produced by the standard solu- 
tion with any given substance must either in itself or by 
an indicator be such, that its termination is unmistakable 
to the eye, and thereby the quantity of the substance with 
which it has combined accurately calculated." ^ 

Single-factor Solutions. — The standard solution may 
be made to any desired strength. If a solution is to be 
frequently used for the determination of a substance, for 
example iron, it is convenient to adjust its strength so 
that one cubic centimeter shows the presence of 0.01, 
0.001 or 0.0001 g. of iron, in which case the determination 
of the weight of iron in grams present in the sample is 
obtained directly by the burette reading. If it is de- 
sired to make up a solution of silver nitrate, one cubic 
centimeter of which is to be equivalent to 0.001 g. of 
chlorine, and the weight of silver nitrate which must be 

1 ''Reaction" might be a better word. 

2 Sutton, " Volumetric Analysis." 

202 



VOLUMETRIC ANALYSIS 203 

contained in a liter of the solution is to be calculated, the 
required weight can be found readily. The reaction is 

M'Cl + AgNOs = M'NOs + AgCl. 

The weight of chlorine which is to be indicated by one 
liter is 

0.001 X 1000. = 1.0000 g. 

The amount of silver nitrate necessary is 
AgNOs _ 169.89 



CI 35.46 



X 1.0000 = 4.7910 g. 



It commonly occurs that a substance with which it is 
desired to make a standard solution cannot be weighed 
out in a state of purity; hence it is necessary to standard- 
ize the solution after it is made up. The methods of 
doing this are many: typical cases will be considered. 

Methods of Standardization. — I. By weighing out 
the amount required and making up to the desired vol- 
ume. This can be done only when the substance is in a 
state of purity and the weighing can be accurately per- 
formed and presents no great difficulties in manipulation. 
Among such substances may be mentioned sodium car- 
bonate, arsenious oxide and sodium oxalate. 

II. By causing a solution to react with a weighed 
amount of a chemically pure substance and measuring 
the volume of the solution necessary to do this. 0.21194 g. 
of anhydrous sodium carbonate require 40.00 cc. of a 
solution of hydrochloric acid to completely neutralize it. 
The strength of the hydrochloric acid per cubic centi- 
meter is 

2HC1 2 (36.47) ^0.21194 ^^^..,. Jrn^ 
Na^Ca = moo- X -4000- = ^'^^^^^^ ^- ^^^ ^^^ ''' 

III. By analyzing gravimetrically a measured portion 
of the solution. If 30.00 cc. of the same solution of 



204 CHEMICAL CALCULATIONS 

hydrochloric acid as above yield 0.4299 g. of silver 
chloride, the content of hydrochloric acid per cubic centi- 
meter is 

HCl 36.47 ,,0.4299 ^^^^.,. „^, 
AiCi = 143:34 >^ 300^ =^ ^-^^^^^^ ^- ^^^ P"^ ''' 

IV. By comparing a measured volume of the solution 
with a measured volume of another solution the strength of 
which is known and with which it reacts. If 45.00 cc. of the 
same hydrochloric acid solution react with 45.20 cc. of 
a solution of sodium hydroxide, each cubic centimeter of 
which contains 0.003982 g. of sodium hydroxide, the 
strength of the hydrochloric acid solution is 

HCl 36.47 ^ 45.20 X 0.003982 „ ^^_ . . ^ „, 

X TFT^ ^ 0.003646 g. HCl per cc. 



NaOH 40.01 45.00 

Equivalent Values of Single-factor Solutions. — A 

solution may be made up to a strength to correspond to 
a simple factor, as, for instance, the solution of silver 
nitrate mentioned above, each cubic centimeter of which 
corresponds to a milligram of chlorine. When such a 
solution is to be used for another determination the value 
of the solution per cubic centimeter must be calculated 
for the new substance. 

Potassium permanganate is a reagent used in a number 
of different volumetric determinations. Its value de- 
pends upon its oxidizing property. Iron is determined 
by it according to the equation 

2 KMn04 + 10 FeS04 + 8 H2SO4 = K2SO4 + 2 MnS04 

+ 5Fe2(S04)3 + 8H20; 

oxalic acid, 

2 KMn04 + 5 H2C2O4 + 3 H2SO4 = K2SO4 + 2 MnS04 
+ IOCO2 + 8H2O; 



VOLUMETRIC ANALYSIS 205 

calcium when in the form of the oxalate, 

2 KMn04 + 5 CaC204 + 8 H2SO4 = K2SO4 + 2 MnS04 
+ 5 CaS04 + 10 CO2 + 8 H2O; 

manganese, 

2 KMn04 + 3 MnS04 + 2 H2O = K2SO4 + 2 H2SO4 

+ 5 Mn02. 

There are many other similar reactions. If one cubic 
centimeter of the potassium permanganate solution repre- 
sents 0.001 g. of iron, then as 2 KMn04 = 10 Fe, and 
2 KMn04 = 5 H2C2O4, it follows that 5 H2C2O4 = 10 Fe: 
the amount of oxalic acid equivalent to one cubic centi- 
meter of this permanganate solution is: 

^iSf '= lorSSf ^ 0.001 =0.000806 g. H2C2O4 per cc. 

In like manner, 2 KMn04 = 10 Fe; 2 KMn04 = 5 Ca; 
then 5 Ca = 10 Fe, and the amount of calcium indicated 
by each cubic centimeter of the permanganate solution is 

jS^ = 10%M) ^ ^'^^^ = 0.0003588 g. Ca per cc. 

Finally, as 2 KMn04 = 10 Fe and 2 KMn04 = 3 Mn, it 
follows that 3 Mn =10 Fe, and the value of the perman- 
ganate solution in terms of manganese is 

3 Mn 3 (54.93) 



10 Fe 10 (55.84) 



X 0.001 = 0.0002951 g. Mn per cc. 



Factor Weights for Volumetric Solutions. — Factor 
weights ^ are as applicable to volumetric as to gravimetric 
analysis. Acetic acid is to be determined by titration with 
sodium hydroxide, one cubic centimeter of which contains 
0.04006 g. What weight of acetic acid should be taken 

1 See p. 176. 



206 CHEMICAL CALCULATIONS 

such that the burette reading in cubic centimeters may 
show the percentage of acetic acid directly? One cubic 
centimeter of the sodium hydroxide solution is equivalent 
to 

?g^ = ~X 0.4006 = 0.06003 g. HC2H3O2 per cc. 

Supposing the acetic acid were 100% pure, the number 
of cubic centimeters required would be 100. This corre- 
sponds to 100 X 0.06003 = 6.003 g. of acetic acid. If 
this weight of acetic acid is weighed out and titrated with 
the sodium hydroxide under consideration, the burette 
reading in cubic centimeters will give the percentage of 
acetic acid present. Thus if 85.00 cc. of sodium hydroxide 
are required for neutralization, the acetic acid contains 
85.00% HC2H3O2 as is readily seen by the following: 

Instead of reading the percentage of acetic acid directly 
from the burette, it may be desirable to adjust the amount 
of acid taken, so that the burette reading multiplied by 
two gives the percentage of acetic acid.^ If half of 6.003 g. 
(= 3.0015 g.), that is, fifty times the factor, is weighed out 
and titrated, the burette reading multiplied by two gives 
the percentage of acetic acid. If 42.50 cc. are required 
for titration, the percentage of acetic acid is 2 X 42.50 = 
85.00%, as is proved: 

^^•%^J,f'^°^ X 100 = 85.00% HC.H3O. 

1 This would be the case if a 50 cc. burette were used, as it would 
obviate the trouble of filling the burette a second time if the per- 
centage exceeds 50%. 



VOLUMETRIC ANALYSIS 207 

To put this in general terms; if the factor weight taken 
is m times the factor, the reading ^ must be multiphed by 
n, m and n being so chosen that their product equals 100. 
Thus, if the factor weight taken is five times the factor, 
the reading must be multiplied by twenty. 

Normal Solutions. — ''A normal solution of a reagent 
is one which contains, in a liter, that proportion of its 
molecular weight in grams which corresponds to one gram 
(gram atom) of available hydrogen or its equivalent." ^ 
Submultiples of the normal solution are used and are 
designated half or seminormal (N/2 or 0.5 N), fifth or 
quintinormal (N/5 or 0.2 N), tenth or decinormal (N/10 
or 0.1 N), hundredth or centinormal (N/100 or 0.01 N), etc. 

According to definition, the following weights of acids, 
bases or radicals contained in a liter are normal solutions: 

Acetic acid (one available H ion) = ^— j = 60.032 g. 

1 The reading may be in cubic centimeters, as in volumetric anal- 
ysis, or in grams or some multiple of the gram, as in gravimetric 
analysis. 

2 Sutton, "Volumetric Analysis." The words "gram atom" in 
parenthesis have been inserted by the author, as in his opinion 
Sutton so intended. As the examples given by Sutton show, the nor- 
mal solution contains 1.008 g. of available hydrogen and not 1.0000 g., 
as his definition would lead one to suppose. His definition would 
be exact if hydrogen were taken as unity in the determination of 
atomic weights. Such, however, is not the case, the basis being 
oxygen = 16 which makes hydrogen = 1.008. 

3 The expression is intended to represent a fraction, the numer- 
ical value of which is obtained by giving to the numerator a number 
equal to the molecular weight of the symbol. For a Hst of acids 
and bases and the values of normal solutions of the same, with the 
indicators to use in each case, see Chem. Ann., pp. 52-53. The 
solubilities of many substances are too small to permit of the prep- 
aration of a solution of normal strength. The values given indicate 
the weight of substance which would be present in a normal solu- 
tion, solubility permitting. Such cases are illustrated by calcium 



208 



CHEMICAL CALCULATIONS 



Ammonia 

Ammonium 
hydroxide 

Hydrochloric 
acid 

Nitric acid 

Barium 
hydroxide 

Calcium 
hydroxide 

Sulphuric acid 

Oxalic acid 

Phosphoric 
acid^ 

Phosphoric 
acid 2 



(one available NH3 ion) = 

[ (one available OH ion) = 

(one available H ion) = 

(one available H ion) = 

(two available OH ions) = 

(two available OH ions) = 

(two available H ions) = 

(two available H ions) = 

(one available H ion) = 

(two available H ions) = 



NH3 
1 

NH4OH 
1 

HCl 
1 

HNO3 

1 

Ba(OH)s 
2 

Ca(OH); 
2 

H2SO4 
2 

C2H2O4 
2 

H3PO4 
1 

H3PO4 



= 17.064 g. 
= 35.08 g. 
= 36.47 g. 
= 63.048 g. 
= 85.71 g. 
= 37.06 g. 
= 49.04 g. 
= 45.01 g. 
= 98.024 g. 
= 49.01 g. 



i 



Following is a list of a few oxidizing agents.^ As one 
oxygen ion is equivalent to two hydrogen ions, the list 
will be evident. 

and barium hydroxides, which on account of their limited solubility 
must be used in fractional normal solutions. Because of its low 
solubility, calcium hydroxide is not used in volumetric operations. 

1 As these solutions are used in neutralization, their value de- 
pends upon the acidity shown. When methyl orange is used as 
the indicator, the same shows color change when one hydrogen is- 
replaced. 

2 When phenol-phthalein is used as the indicator, the color change 
takes place when two hydrogens are replaced. 

3 For a list of volumetric oxidizing and reducing agents with the 
values of a normal in each case, see Chem. Ann., pp. 54-55. 

For a list of precipitation agents and data on the strengths of 
normal solutions of the same, see Chem. Ann., p. 56. 



VOLUMETRIC ANALYSIS 209 

1 Iodine (liberates or is = 1 H ion) = - = 127.92 g. 

1 Barium peroxide (liberates 1 O = 2 H ions) = — ~ = 84.7 g. 

1 Potassium di-) .,., , „^ „ tt . . K2Cr207 .„ „„ 

, ^ [ (liberates 3 O = 6 H ions) = — = 49.08 g. 

chromate ) ^6 

1 Potassium per- ) .,., , ^, ^ e tt • m KMnOi oi ^o 

, ^ [ (liberates 2| O = 5 H ions)i = — = 31.63 g. 

manganate ) 5 

1 Arsenious oxide (takes up 2 O = 4 H ions) = — '^— = 49.5 g. 

In order to determine the amount of substance which 

must be present to form a solution of a required normality, 

the particular reaction in which the solution takes part 

must be known. Thus, in an acid solution, two molecules 

of potassium permanganate liberate five atoms of oxygen. 

Permanganic acid is HMn04; the anhydride would be 

2 HMn04 = MnaOy + H2O. 

In acid solution, the decomposition may be abbreviated to 

MnsOy = 2 MnO + 5 0. 

Then as 5 O = 10 H, the weight of salt requisite in a liter 

of a normal solution when used in an acid condition is 

2 KMn04 KMn04 158.15 ^^ .„ ^, ^ ^ 
j^ = — ^ — = — g — = 31.63 g. KMn04. 

In an alkaline solution the essential features may be repre- 
sented by 

2 MnaOT = 2 MnOs + 3 0. 

Hence, when potassium permanganate is to be used in an 
alkaline solution, the weight necessary for a liter of a nor- 
mal solution is 

2KMr^. ^ mo. ^ 15|1_5 ^ ,^,,,,^ kM„0..^ 

1 That is, two molecules of KMn04 yield five oxygens in an acid 
solution. 

2 Potassium permanganate is rarely titrated in an alkaline or 
neutral solution. An example is the titration of manganese. See 



210 CHEMICAL CALCULATIONS 

The calculation of the normal quantity of a salt is not 
always such a simple matter. The normal quantity of 
hydrochloric acid being 36.47, in the reaction 
NH4OH + HCl = NH4CI + H2O, 

the weight of ammonium chloride equivalent to this 
weight of hydrochloric acid, and, consequently, the normal 
quantity of ammonium chloride in this reaction, is 

■S?- if =<».„ = 53...,. 

In like manner the amount of ammonium sulphate is 
obtained from the equation 

2 NH4OH + H2SO4 = (NH4)2S04 + 2 H2O; 

consequently, the normal weight (the weight in grams per 
liter, the normal weight of sulphuric acid being 49.04 g.) 
is 

(NH4)2 S04 132.15 ^ ,„ „, _ _„ 

Simplification of Calculations by the Use of Normal 
Solutions. — Knowing the reaction, when normal quan- 
tities are used, the weights of the substances involved may 
be found by inspection. Defining oxidation as the process 
of adding oxygen to, or the removal of hydrogen from, 
or raising the valence of, the substance oxidized, and 
reduction as the reverse, the normal weights of com- 
pounds that are neither acids nor bases, but will react 
with oxidizing or reducing compounds, are readily deter- 
mined. The oxidation of iron raises its valence from two 
to three, a gain of one in valence, which is equivalent to 

reaction, Prob. 391. For other examples, see phosphoric acid with 
different indicators, p. 208. When a normal solution of potassium 
permanganate is spoken of, the solution equivalent to five oxygens 
is understood unless specifically qualified. 



VOLUMETRIC ANALYSIS 211 

one hydrogen. Hence, the normal amount of iron, i.e., 
the amount of iron shown by a Hter of any normal oxidiz- 
ing or reducing reagent used for its determination, is 

55 84 

— ij — = 55.84 g. Antimonious salts when acted upon by 

certain oxidizing agents are converted into antimonic 
salts, i.e., antimony is changed from the trivalent to the 
pentavalent condition, a gain in valence of two; hence, 
the weight of antimony shown by a normal weight of 
oxidizing or reducing substances is 

Sb 120.2 „„^„ 
_=-__ = 60.10 g. 

Normal solutions are of such strength that equal 
volumes are exactly equivalent; thus, one normal cubic 
centimeter of an acid exactly neutralizes one normal 
cubic centimeter of a base, producing a neutral solution.^ 
Knowing the normality of a solution, its value in terms of 
any constituent with which it reacts can be calculated 
easily, and can often be done by inspection. If 30.00 cc. 
of an N/10 solution of sodium hydroxide are used to 
determine hydrochloric acid in a 2.0000 g. sample, the 
amount of hydrochloric acid present is (1.00 cc. N/10 HCl 

= lOOOxTO = 0-003647 g. HCl) 

30.00 X 0.003647 = 0.10941 g.; 

1 There is another system of concentrations. This is the Molar 
system and is used chiefly in physical chemistry. A molar solution 
contains one gram molecule of solute per liter of solution. Under 
this definition a molar solution of hydrochloric acid contains 36.468 
g. HCl per liter as does also a normal solution. A molar solution 
of sulphuric acid contains 98.086 g. H2SO4 per liter, while a normal 
solution contains 49.043 g. So it is seen that a molar and a normal 
solution may or may not be equivalent. 



212 CHEMICAL CALCULATIONS 

and as 2.0000 g. of sample were taken, the percentage of 

hydrochloric acid is 

30.00 X 0.003647 ^ , 
2:0000 ■ ^ ^^^ = ^'^^^' ^^^' 

If the 30.00 cc. had been used in the determination of 

sulphuric acid, the same amount being taken (1.00 cc. 

1 XT O/^ 

N/10 H2SO4 = iqqqO{q = 0.004904 g. H2SO4), the result 
would have been 

^°-°%>^,Sg^^^"^ X 100 = 7.36% H.SO. 

Again, potassium permanganate is used to determine 
iron, calcium in the form of oxalate, and also nitrous acid. 
45.00 cc. of a 0.1 N^ solution of potassium permanganate 
are required to bring about the reactions. The amounts 
present can be readily determined without recourse to 
stoichiometric ratios in the form of a reaction. Iron is 
oxidized by permanganate, the valence being increased 
by one. If all the iron were present in the ferrous con- 
dition, then as the amount of iron shown by one cubic 
centimeter of 0.1 N potassium permanganate is 

i-xWxl6 = 0-00^^«*^-' 

the total amount of iron is 

45.00 X 0.005584 = 0.2513 g. Fe. 
The calcium is separated as the oxalate which on treat- 
ment with sulphuric acid liberates one molecule of oxalic 
acid for each atom of calcium present. The oxalic acid 
is oxidized by the permanganate, one atom of oxygen 
being taken up for each molecule of oxalic acid present. 
Then, as one oxygen is equivalent to two hydrogens, the 

^ The normality of a solution may as well be expressed as a 
decimal as a fraction; in fact, the decimal system is to be preferred. 



VOLUMETRIC ANALYSIS 213 

amount of calcium equivalent to a cubic centimeter of a 
tenth normal oxidizing solution is 

The amount of calcium present is 

45.00 X 0.0020035 - 0.090158 g. Ca. 

Potassium permanganate oxidizes nitrous acid to nitric 
acid, a gain of one atom of oxygen which is equivalent 
to two hydrogens. The amount of nitrous acid equivalent 
to a cubic centimeter of a decinormal solution is 
HNOo 
2 X 1000 'x 10 - 0-0°^^^^ g- 

The amount of nitrous acid present is 

45.00 X 0.002351 - 0.1058 g. HNO2. 

Calculation of Normality. Factor to a Given Nor- 
mality. — Thus far solutions have been considered as 
being adjusted to the normality indicated. This is a 
matter of considerable difficulty and it is a general prac- 
tice to calculate the strength of solutions by methods 
already given, not attempting to have the normality more 
than approximate, the exact strength, however, always 
being known. A solution shows on standardization 
0.02458 g. of sulphuric acid per cubic-centimeter. A nor- 
mal solution of this acid contains 0.049045 g. per cubic 
centimeter; then the normality of this solution is 

0;^ = 0.5012 N. 
0.049045 

This shows the solution to be very nearly half normal, 
being a little strong. The factor to exactly half normal 
is 

0.5012 



0.5000 



= 1.002. 



214 CHEMICAL CALCULATIONS 

The use of this factor is to calculate the number of cubic 
centimeters used in an analysis to an equivalent number 
of cubic centimeters of exactly half normal, that the 
amount of substance indicated by it may be readily deter- 
mined as previously given. For example, 30.25 cc. of 
this sulphuric acid are used in determining sodium hydrox- 
ide. The number of cubic centimeters of exactly N/2 
sulphuric acid equivalent to this number of cubic centi- 
meters is 30.25 X 1.002 = 30.31 cc. The amount of 
sodium hydroxide present is 

40 01^ 

X 30.31 = 0.6064 g. NaOH. 



2 X 1000 



It is often desirable to know the number of cubic 
centimeters of a solution of given normality equivalent 
to a given number of cubic centimeters of another solu- 
tion of different normality. This number is inversely 
proportional to the normality of the solutions under 
consideration; the greater the normality, the fewer the 
number of cubic centimeters necessary to bring about 
the reaction. It is at once apparent that it will require 
twice as many cubic centimeters of a half normal solution 
(N/2 or 0.5 N) to accomplish the same amount of chemical 
change as with a normal solution (N, N/1 or 1 N). Nor- 
malities expressed in the decimal system (0.5 N, 1 N) 
readily lend themselves to calculations of this nature. 

1 With a little practice and familiarity, the use of solutions stand- 
ardized in the normal system will become so easy that in large part 
the calculations may be performed mentally. An operator knows 
that a normal solution of any acid indicates 0.04001 grams of sodium 
hydroxide per cubic centimeter, knowing the molecular weight of 
sodium hydroxide to be 40.01, and a half normal solution being of 
half the value of a normal solution, the value 0.020005 g. of sodium 
hydroxide indicated by an N/2 solution of any acid is obtained 
by inspection. 



VOLUMETRIC ANALYSIS 215 

Thus, how many cubic centimeters of a 0.1045 N solution 
are equivalent to 35.00 cc. of a 0.1003 N solution? The 
number of cubic centimeters required being inversely 
proportional to the normalities or strengths of the solu- 
tions gives 

^4^ X 35.00 = 33.59 cc. of 0.1045 N sol. = 35.00 cc. 
0.1045 

of 0.1003 N sol. 
This may be expressed in algebraic form. Let 

V = number of cubic centimeters requisite for a cer- 
tain reaction, 

N' = normality of the solution corresponding to V 
cubic centimeters, 

V = number of cubic centimeters of a different 
strength requisite for the same reaction, 

N" = normality of the solution corresponding to V" 
cubic centimeters. 

Then 

Taking the same example; substituting gives 

V X 0.1045 = 35.00 X 0.1003, 

V = ^4^ X 35.00 = 33.59 cc. 

0.1045 

To state another problem bearing on this same subject. 
11.00 cc. of a 0.5200 N solution of hydrochloric acid are 
required to neutralize 50.00 cc. of a solution of barium 
hydroxide. What is the normality of the barium hydrox- 
ide solution? The normality being inversely proportional 
to the number of cubic centimeters requisite: 



216 CHEMICAL CALCULATIONS 

|i^ X 0.5200 = 0.1 144 N, 

or substituting in the formula gives 

50.00 X N' = 11.00 X 0.5200, 

N' = ^^ X 0.5200 = 0.1144 N. 

Again, how many cubic centimeters of a solution whose 
factor to N/10 is 0.9924 are equivalent to 18.00 cc. of a 
solution whose factor to N/2 is 1.005? The number of 
cubic centimeters of N/10 solution equivalent to a solu- 
tion whose factor to N/10 is 1.005, is 18.00 X 1.005 = 
18.09 cc. This number of cubic centimeters of N/10 
solution is equivalent to 18.09 X 5 = 90.45 cc. of N/2 
solution. Then the number of cubic centimeters of a 
solution whose factor to N/2 is 0.9924 equivalent to 90.45 
cc. N/2 solution is 

Solving in another manner: The normality of the first 
solution, expressed decimally, is 0.1 X 0.9924 = 0.09924 N. 
The normality of the second solution, similarly expressed, 
is 0.5 X 1.005 = 0.5025 N. Then 
0.5025 



0.09924 



X 18.00 = 91.14 cc. 



Volumetric Determinations Using Two Solutions. 
Back Titrations. — Many volumetric analyses are car-, 
ried out by the use of two solutions. These methods 
generally consist in adding an excess of the active reagent 
and determining this excess by means of a suitable 
second solution with which the first reacts. It is nec- 
essary to know the relative strengths of the two solutions 
employed so that the amount of the solution used in pro- 



VOLUMETRIC ANALYSIS 217 

ducing the desired reaction may be ascertained. When 
the two solutions are exactly equivalent, cubic centimeter 
to cubic centimeter, subtraction of the volumes used gives 
the amount of the first solution necessary to bring about 
the desired reaction. 

A sample of calcium carbonate is to be analyzed, 
phenol-phthalein being used as the indicator, which is 
affected by the carbonic acid liberated when the calcium 
carbonate is treated with an acid; hence, the end point 
in this titration would occur too soon, were it attempted 
to titrate directly to a finish with sulphuric acid alone. 
A solution of sulphuric acid of exactly 0.1 N strength and 
a solution of exactly 0.1 N sodium hydroxide are em- 
ployed. 0.2000 g. of the sample are weighed out, water 
added, and treated with 45.00 cc. of the 0.1 N sulphuric 
acid and the solution boiled to expel the carbon dioxide. 
The solution is now acid with sulphuric acid, 6.45 cc. 
of 0.1 N sodium hydroxide being required to produce 
neutrality. The amount of sulphuric acid used in de- 
composing the calcium carbonate is evidently 45.00 — 
6.45 = 38.55 cc. of 0.1 N solution. Assuming that there 
are no other carbonates present, the equivalent of one cubic 
centimeter of 0.1 N solution of sulphuric acid in terms of 

calcium carbonate being ^ = 0.0050035 g., 

the percentage of calcium carbonate in the sample taken is 

38.55 X 0.0050035 



0.2000 



X 100 = 96.44% CaCOa. 



Calcium acetate, Ca(C2H302)2 • H2O, is to be examined 
for the percentage of the same present in a sample weigh- 
ing 3.7246 g. The calcium acetate is digested with an 
excess of syrupy phosphoric acid, the acetic acid liberated 
being distilled into 50.00 cc. of 1.005 N sodium hydroxide 



218 CHEMICAL CALCULATIONS 

solution. The excess of sodium hydroxide required 
55.25 cc. of 0.2012 N sulphuric acid for neutralization. 

50.00 X 1.005 = 50.25 cc. of 1 N NaOH solution taken, 

^1^^ X 55.25 = 55.58 cc. of 0.2 N H2SO4 for back 

titration, 

55.58 cc. of 0.2 N H2SO4 = 0.2 X 55.58 = 11.12 cc. of 
1 N H2SO4, 

50.25 - 11.12 = 39.13 cc. of 1 NNaOH consumed. 

Then, as Ca(C2H302)2 • H2O ^ 2 HC2H3O2 = 2 NaOH, the 
amount of calcium acetate indicated by one cubic centi- 
meter of normal sodium hydroxide being 

Ca(C2H302)2 » H2O ^ ^^^__ 
2^0600 = 0-088065 g., 

the per cent of calcium acetate is 
39.13 X 0.088065 



3.7246 



X 100 = 92.52% Ca(C2H302)2 • H2O. 



Titration of ** Mixed Acid." — A mixture of sub- 
stances may be analyzed volumetrically. A sample of 
*' mixed acid" contains sulphuric acid, nitric acid and 
lower oxides of nitrogen, chiefly N2O3, which result from 
mixing the two acids. 

A sample is taken, treated with water and evaporated 
on the steam bath, the operation being repeated twice, 
which treatment leaves only sulphuric acid which is 
titrated with standard alkali solution. A second sample 
is titrated with standard alkali for total acidity and a 
third sample with potassium permanganate for nitrous 
anhydride. 



VOLUMETRIC ANALYSIS 219 

Data for determination of mixed acid : 
For total acid : 

Weight of sample taken = 4.6514 g. 
NaOH to neutralize = 74.10 cc. 

1.00 cc. NaOH sol. ^ 0.052109 g. H2SO4. 

Per cent total acid as H2SO4 is 

ZM^I^Of^iOix 100^ 83.01%. 

For sulphuric acid (HNO3 + HNO2 being driven off by 
evaporation) : 

Weight of acid taken = 8.8872 g. 
NaOH to neutralize = 72.20 cc. 
1.00 cc. NaOH sol. = 0.052109 g. H2SO4. 

Per cent sulphuric acid is 

For nitrous anhydride : 

Weight of sample taken = 17.000 g. 
0.1 NKMn04 to oxidize 

N2O3 to N2O5 = 8.60 cc. 

1.00 cc. of 0.1 N KMn04 ^ 0.001900 g. N2O3. 

Per cent nitrous anhydride is 

?:55X^ = 0.096%= 0.10%. 

To calculate the composition of the mixed acid: 

83.01 - 42.33 - 40.68% HNO3 + HNO2 as H2SO4. 

The amount of acidity as nitric acid is 
2 HNO3 2 (63.02) 



H2SO4 98.09 

as HNO3. 



X 40.68 = 52.27% HNO3 + HNO2 



220 CHEMICAL CALCULATIONS 

The equivalent of N2O3 in HNO3 is 

2 HNO3 2 (63.02) 

"N^OT = "T6:0^ X 0-09^ = 0.16%. 
The amount of nitric acid present is 

52.27 - 0.16 = 52.11% HNO3. 
From these figures the analysis of the mixed acid is 

Sulphuric acid 42 33 

Nitric acid " 52 n 

Nitrogen trioxide q ' 2q 

Water (by difference) ' * ' 5 ' 4^ 

Too. 00 
Volumetric Analysis Using Two Indicators. — Mix- 
tures may also be analyzed by taking advantage of the 
conduct of different salts toward indicators.^ For exam- 
ple, phenol-phthalein reacts alkahne to sodium carbonate, 
neutral to bicarbonate, and acid to carbonic acid.^ If 
a mixture of sodium hydroxide and sodium carbonate is 
treated with a known amount of hydrochloric acid till 
the solution reacts just acid to phenol-phthalein, the 
following will be the reaction: 

(1) X NaOH + y Na2C03 + (x + y) HCl = y NaHCOs 

-i-xR,0+ (x + 2/)NaCl. 
If the titration is continued, boiling out the carbonic acid, 
the solution will react neutral to methyl orange when the 
bicarbonate is decomposed : 

(2) y NaHC03 + yRC\ = y NaCl + yR,0 + y CO2. 
* See Chem. Ann., p. 51. 

2 Phenol-phthalein as an indicator is used to show the presence 
of minute amounts of strong bases or weak acids. Methyl orange 
indicates the presence of minute amounts of weak bases and strong 
acids. For strong acids and strong bases either will serve. Methyl 
red is an indicator which acts similarly to methyl orange and pos- 
sesses advantages over the latter, being nearly colorless in a basic 
and a strong red in an acid solution. 



VOLUMETRIC ANALYSIS 221 

Then, if p represents the total hydrochloric acid used, and 
q the hydrochloric acid used to convert the sodium hydrox- 
ide and the sodium carbonate into sodium chloride and 
sodium bicarbonate, 

x-\-2y = p 

x-\r y = q 

y = p-q 

X = 2q — p. 

To illustrate: one-half gram of a sample composed of 
sodium hydroxide and sodium carbonate requires 62.50 cc. 
N/10 hydrochloric acid to neutral reaction with phenol- 
phthalein. The titration is continued; 100.00 cc. of the 
same acid are required for complete neutrality (methyl 
orange) . The last addition of hydrochloric acid is a meas- 
ure of the sodium carbonate present, for all the sodium 
hydroxide was neutralized with the first addition of acid. 

100.00 - 62.50 = 37.50 cc. of N/10 HCl for NaHCOs 
->NaCl, H2O, CO2. 

The amount of bicarbonate is equivalent to the amount of 
sodium carbonate originally present. 1 cc. of N/10 HCl = 

^ = 0.010600 g. of sodium carbonate, as only 

iUUU /\ -LU 

one acid hydrogen is present in the acid carbonate. Then 

37.50X0.010600^^ ,^„ .rn rnoy at nr, 
05000 ^ ^^^ = 79.50% NaaCOs. 

The same amount of hydrochloric acid is used to convert 
the bicarbonate into sodium chloride, carbon dioxide and 
water as was required to convert the sodium carbonate 
into bicarbonate. Therefore, 62.50 - 37.50 = 25.00 cc. 
of N/10 HCl, used in reacting with the sodium hydroxide. 



222 CHEMICAL CALCULATIONS 

1 cc. of N/10 HCl ^ 1^^^^ = 0.004001 g. sodium hy- 
droxide. Then 

25:00^0^ X 100 = 20.01% NaOH. ' 

Algebraically, according to the equations given, letting 
X = cc. of N/10 acid combining with NaOH, 
y = cc. of N/10 acid required to change Na2C03 to 
NaHCOs and NaHCOs to NaCl, CO2 and H2O. 
Then 

X = 125.00 - 100.00 = 25.00 cc. N/10 HCl, 
y = 100.00 - 62.50 = 37.50 cc. N/10 HCl. 
The calculation is completed as given above. 

Volumetric Titration of Oleum. — When an oleum 
contains free sulphurous anhydride, an interesting and 
important case of indirect volumetric analysis results. 
Such an oleum contains sulphuric acid, sulphurous an- 
hydride (SO2) and sulphuric anhydride (SOs).^ A weighed 
sample is dissolved in water and titrated with a standard 
alkali solution, when all the constituents are acted upon. 
Thus, for sulphuric acid: 

H2SO4 + 2 NaOH = Na2S04 + 2 H2O. 

The sulphuric anhydride dissolves in water to form sul- 
phuric acid and is titrated: 

SO3 + H2O = H2SO4, 
H2SO4 + 2 NaOH = Na2S04 + 2 H2O. 
Likewise, the sulphurous anhydride dissolves: 
SO2 + H2O = H2SO3. 

1 There may be other impurities, such as solid particles, etc., 
but for these calculations, only the three constituents enumerated 
will be considered as being present. The method is easily extended 
to cover impurities. 



VOLUMETRIC ANALYSIS 223 

When phenol-phthalein is used as an indicator, the reac- 
tion with the alkah is 

H2SO3 + 2 NaOH = NasSOa + 2 H2O, 

while with sodium hydroxide, using methyl orange as an 
indicator, it is 

H2SO3 + NaOH = NaHSOa + H2O. 

The sulphurous anhydride is estimated in a separate 
sample by dissolving the oleum in water and titrating 
against an iodine solution: 

H2SO3 + H2O + I2 = H2SO4 + 2 HI. 

An example of an oleum analysis follows: 5.0000 g. 
of an oleum are dissolved in water and the volume made 
up to 500 cc. Of this solution, 100 cc. (= 1.0000 g. of 
sample) are taken and titrated with N/10 iodine solution, 
7.80 cc. being required. A similar portion is taken 
and titrated with N/5 sodium hydroxide, using phenol- 
phthalein as an indicator, 122.81 cc. being required. To 
calculate the composition of the oleum: 

7.80X0.0032035 ,,,^^ or.no/Qn 
fOOOO ^ 2.50% SO2, 

122.81 N/5 sol. = 2 X 122.81 = 245.62 cc. N/10 sol., 
245.62 - 7.80 = 237.82 cc. N/10 NaOH 

required for the titration of the sulphuric acid and sulphur 
trioxide.^ 

^ The 7.80 cc. are subtracted, this being the number of cubic 
centimeters of N/10 solution of sodium hydroxide used in neutral- 
izing the sulphurous acid. If methyl orange had been used, 253.42 
cc. of N/10 sodium hydroxide would have been necessary for the 
total acidity titration, but 15.6 cc. of N/10 solution would have to 
be deducted from the 253.42 cc. N/10 sodium hydroxide required 
for the total acidity, leaving 237.82 cc. necessary for the sulphuric 
acid and the sulphuric anhydride, as before. 



224 CHEMICAL CALCULATIONS 

237.82X0.0040035 ,,^.. q^ 910/ w ion 
TOOOO ^ 95.21% total SO3, 

95.21 + 2.50 = 97.71% SO3 + SO2, 
100.00 - 97.71 = 2.29% H2O. 

Calculating this percentage of water into equivalent sul- 
phuric acid: 

jg^QjgX 2.29 = 12.47% H2SO4. 
Then 

100.00 - (12.47 + 2.50) = 85.03% free SO3. 
To summarize, the oleum is composed of 

Per cent 

H2SO4 12.47 

SO3 85.03 

SO2 2.50 

To calculate algebraically, let 
X = per cent of H2SO4, 
y = per cent of SO3, 
z = per cent of SO2, 

A = percentage total acidity as H2SO4, not including 
that due to SO2, 

Then 

x + y-{-z = 100 

x-i- y =100-2 

X = 100 - (2 + y). 

From the conditions of the problem: 
X+ y =100- z 
x+fy = A 

_ A+z- 100 ^ A +z- 100 
^ / - 1 0.22505 

= 4.4436 {A + z- 100). 



VOLUMETRIC ANALYSIS 225 

Solving the given problem by this method: 
z — 2.50 (as before), 

. 237.82X0.0049045 ,,,^^ h^a.o/ 
" LOOOO ^ "" 116.64%, 

y = 4.4436 (116.64 + 2.50 - 100) = 4.4436 X 19.14 

= 85.05%, 
X = 100 - (2.50 + 85.05) = 12.45%, 

which results are the same as were obtained by the method 
given before.^ 

Adjustment of the Strength of Solutions. — Having a 
solution containing 47% HNO3, and a weaker solution of 
23% HNO3, let it be required to prepare, by mixing these 
two solutions, 200 kg. of a solution which will contain 
39% HNO3. To find the weights of the two solutions, 
which, when mixed, will give the desired quantity of acid 
of the specified strength, let y = the weight of the 
weaker acid (23% HNO3) to be taken; then 200 - y = 
the weight of the stronger acid (47% HNO3) to be taken. 
Algebraically : 

0.23 y + 0.47 (200 - y) = 0.39 (200), 

y = 66.67 kg. 23% HNO3, 
200 - 66.67 = 133.33 kg. 47% HNO3. 

Using this method, a general formula may be obtained. 
Let 

A = strength of stronger solution, 
B = strength of weaker solution, 
D = strength of desired solution, 
X = amount of stronger solution to be taken, 
y = amount of weaker solution to be taken, 
z = amount of solution desired. 

* An oleum analyzed by these methods must necessarily foot 
up to 100% as the free SO3 is calculated by difference. 



226 CHEMICAL CALCULATIONS 

Then 



By + Aiz- 


-y) = 


Dz, 


By + Az- 


Ay = 


Dz, 


y{B-A)- 


= z{D 


-A), 


D-A 


A 


-D 


y^B-A 


' = A 


-b'' 


x + y = z, 






X = z - 


-y- 





Again, suppose the problem is to alter the strength of a 
definite quantity of acid. Given, an acid of 47% HNO3, 
let it be required to dilute 200 kg. of this acid to 39% 
HNO3, using 23%HN03. What weight of 23% HNO3 
must be added to 200 kg. to make the mixture 39% HNO3? 
Using the same symbols as before : 

0.47 (200) + 0.23 y = 0.39 (200 + y), 

y = 100 kg. 23% HNO3. 
In general terms 

Ax + By = D{x + y), 
y{B-D) = x{D- A), 
_ D-A _ A-D 
y - B-D^~ D-B^' 

The following formulas, derived by the processes given 
above, are very convenient for adjusting the strength of 
solutions. In these, let 

A = actual concentration of the solution that is to be- 

corrected, 
D = desired concentration, 
B = concentration of diluting solution, 
C = concentration of strengthening solution, 
X = amount of stronger solution to be added, taken 

or prepared, 



VOLUMETRIC ANALYSIS 227 

y = amount of weaker solution or water to be added 

or taken, 
z = amount of solution desired or given. 

Formulas (a) are to be used when a definite amount of 
the solution is to be prepared. Formulas (b) are to be 
used when a definite amount of solution is to be corrected^ 
i.e., strengthened or weakened. 

I. To dilute a solution with water: ^ 

{a) -^z = X, y = z- X. 

(b) ^ x = y. 

II. To dilute a strong liquid with a weaker liquid: 

, . A-D 

(^) A - B ^ ^ ^' X = z-y. 

1 The Giles flask is a convenient instrument for preparing a 
solution of definite strength. The neck of the flask is blown out 
into a bulb. The lower mark below the bulb is the mark for the 
indicated capacity of the flask, while above the bulb is another 
mark indicating a capacity between it and the lower mark of one- 
tenth the indicated capacity of the flask. To make up a standard 
solution, 11/10 plus a small amount required for the volume indicated 
are weighed out and introduced into the flask. Water is filled to the 
upper mark and a sample pipetted out of the flask, such sample 
not exceeding the capacity of the bulb between the upper and lower 
marks. This is titrated and the solution drawn off to the lower 
mark (the mark of the indicated capacity). As the solution was 
purposely made a trifle strong, water must be added to adjust to 
the strength desired. The amount of liquid being known, the 
amount of water to add is calculated by (6). 



228 CHEMICAL CALCULATIONS 

III. To strengthen a weak solution with a stronger solution: 
W Q _ ^ z = x, y = z-x. 



PROBLEMS 

390. How many grams per liter must a solution of potassium 
y dichromate contain so that each cubic centimeter shall show 

0.005000 g. of iron according to the reaction : 

6 FeCl2+K2Cr207+14 HC1 = 6 FeCl3+2 KCl+2 CrCl3+7 H2O? 

Ans. 4.3905 g. 

391. How much potassium permanganate per liter must a 
solution contain to be of such strength that each cubic centi- 

> meter is equivalent to 0.001000 g. of manganese? The reaction 
is 

2 KMn04 + 3 MnS04 + 7 H2O = 2 KHSO4 + H2SO4 + 5 HaMnOa. 

Ans. 1.9180 g. 

392. It is desired to make up a solution of potassium 
dichromate of such strength that one cubic centimeter will indi- 

■^ cate 0.001000 g. of iron, (a) How many grams of the potas- 
sium dichromate must be contained in a liter? (6) How many 

1 These formulas are universally accurate when the weights of 
solution are considered. If the solutions are dilute, the specific 
gravities may be considered, with but little error, as being the same 
as water. In this case, the formulas may be used for volumes. 
When this assumption is not permissible, the weights may be cal- 
culated, and knowing the specific gravities of the components, the 

volumes requisite calculated from the formula, volume = 

sp. gr. 

On mixing such concentrated solutions, to use these formulas it 

must be assumed that the volumes are additive, i.e., no change of 

volume takes place. 



VOLUMETRIC ANALYSIS 229 

grams of potassium dichromate, that each cubic centimeter shall 

show 0.001000 g. FesOs? The reaction is 

6 FeCl2+K2Cr207 + 14 HCl = 6 FeCh + 2 KCl + 2 OrClg + 7 H2O. 

Ans. (a) 0.8781 g. 
(b) 0.6141 g. 

393. How many grams of potassium permanganate per liter 
must a solution contain, each cubic centimeter of which is to be 
equivalent to 0.001000 g. of iron according to the equation: 

2 KMn04 + 10 FeS04 + 8 H2SO4 = K2SO4 + 2 MnS04 
+ 5Fe2(S04)3 + 8H20. 

Ans. 0.5660 g. 

394. Calcium is separated as the oxalate which is treated with 
sulphuric acid and the oxalic acid liberated is titrated with 
potassium permanganate. The reactions are 

CaCl2 + (NH4)2C204 = CaC204 + 2 NH4CI, 
CaC204 + H2SO4 = CaS04 + H2C2O4, 
5 H2C2O4 + 2 KMn04 + 3 H2SO4 = K2SO4 + 2 MnS04 
+ 10 CO2 + 8 H2O. 

How many grams per liter of potassium permanganate are re- 
quired that one cubic centimeter shall show (a) 0.001000 g. of 
calcium? (b) 0.001000 g. of calcium oxide? (c) 0.01000 g. 
of CaS04 • 2 H2O? Ans. (a) 1.5776 g. 

(6) 1.1300 g. 

(c) 3.6715 g. 

395. What weight of silver nitrate must be taken, dissolved 
and made up to a liter that each cubic centimeter shall be equiva- 
lent to a milligram of sodium chloride according to the reaction: 

NaCl + AgNOa = NaNOa + AgCl. 

Ans. 2.9060 g. 

396. Arsenious oxide acts as a reducing agent: 

AS2O3 + 20 = AS2O5. 

How much arsenious oxide must be weighed out, dissolved and 
made up to a liter that each cubic centimeter shall take up 
0.001000 g. of oxygen or its equivalent? Ans. 6.1850 g. 



230 CHEMICAL CALCULATIONS 

397. A solution contains 67.0000 g. of sodium oxalate in a 
liter. If it reacts as follows: 

Na2C204 + H2SO4 = Na2S04 + H2C2O4, 
H2C2O4 = H2O + CO2 + CO, 
CO + = CO2, 

how much oxygen is taken up per cubic centimeter? 

Ans. 0.0080 g. 

398. A solution contains 10.0000 g. of silver nitrate made up 
to a Kter. What is the content of silver per cubic centimeter? 

Ans. 0.006350 g. 

399. 30.20 cc. of a solution of sulphuric acid neutralizes 
0.1708 g. of pure anhydrous sodium carbonate. What is the 
amount of sulphuric acid in tlie solution per cubic centimeter? 

Ans. 0.0052335 g. 

400. 0.5000 g. of pure arsenious oxide are dissolved: 100.00 
cc. of a solution of iodine are necessary to oxidize the arsenious 
oxide: 

AS2O3 + 2 12 + 2 H2O = AS2O5 + 4 HI. 

What is the strength of the iodine solution per cubic centimeter? 

Ans. 0.012826 g. 

401. 0.1062 g. of iron 99.8% pure are converted into ferrous 
sulphate and require 19.00 cc. of potassium permanganate to 
oxidize to ferric sulphate: 

2 KMn04 + 10 FeS04 + 8 H2SO4 = K2SO4 + 2 MnS04 
+ 5 Fe2(S04)3 + 8 H2O. 

What is the content of potassium permanganate per cubic centi- 
meter in this solution? Ans. 0.003157 g. . 

402. 0.2680 g. of pure sodium oxalate are weighed out and 
dissolved in water. 40.00 cc. of potassium permanganate 
solution are required to react wath this amount of sodium oxa- 
late: 

2 KMn04 + 5 Na2C204 + 8 H2SO4 = K2SO4 + 2 MnS04 
+ 5 Na2S04 + 10 CO2 + 8 H2O. 



VOLUMETRIC ANALYSIS 231 

What is the content of potassium permanganate per cubic 
centimeter? Ans. 0.0031606 g. 

403. 50.00 cc. of a solution of hydrochloric acid are treated 
with silver nitrate, yielding 0.7167 g. of silver chloride. What 
is the weight of hydrochloric acid per cubic centimeter? 

Ans. 0.003647 g. 

404. A solution of sulphuric acid is standardized by precipi- 
tation as barium sulphate. 30.00 cc. gave 0.3500 g. of barium 
sulphate. What is the strength of the sulphuric acid solution 
per cubic centimeter? Ans. 0.004903 g. 

405. 45.00 cc. of a solution of barium hydrate when precipi- 
tated with sulphuric acid yield 0.6842 g. of barium sulphate. 
(a) What is the content of barium hydroxide per cubic centi- 
meter? (6) To how many grams of hydrochloric acid per cubic 
centimeter is this solution equivalent? Ans. (a) 0.011164 g. 

(6) 0.004751 g. 

406. 25.00 cc. of a solution of sulphuric acid yield 0.5852 g. 
of barium sulphate. What is the content of sulphuric acid per 
cubic centimeter of this solution? Ans. 0.0098355 g. 

407. What is the strength of a solution of barium hydroxide 
if 30.00 cc. require for neutrahzation 59.80 cc. of a solution of 
sulphuric acid known to contain 0.004900 g. per cubic centi- 
meter? Ans. 0.01707 g. 

408. It is found that 32.50 cc. of a solution of sodium hydrox- 
ide exactly neutralize 30.00 cc. of a solution of sulphuric acid 
known to contain 0.04902 g. per cubic centimeter. What is the 
strength of the sodium hydroxide solution per cubic centimeter? 

Ans. 0.03691 g. 

409. A solution of sodium hydroxide is equivalent to 0.004875 
g. of sulphuric acid per cubic centimeter, (a) What is its equiv- 
alent in nitric acid per cubic centimeter? (h) Hydrochloric 
acid? Ans. (a) 0.006264 g. 

(6) 0.003625 g. 



232 CHEMICAL CALCULATIONS 

410. One cubic centimeter of a solution of potassium perman- 
ganate is equivalent to 0.005000 g. of iron. What is its equiv- 
alent of KH3(C204)2 • 2 H2O, the reaction being 

10 KH3(C204)2 + 8 KMn04 + 17 H2SO4 = 9 K2SO4 
+ 8 MnS04 + 40 CO2 + 32 H2O? 

Ans. 0.005689 g. 

411. One cubic centimeter of a solution of potassium perman- 
ganate is equivalent to 0.005585 g. of iron. To how much 
potassium ferrocyanide, K4Fe(CN)6 • 3 H2O, per cubic centi- 
meter, is this solution equivalent, the reaction being 

10 K4Fe(CN)6 + 2 KMn04 + 8 H2SO4 = 10 K3Fe(CN)6 
+ 6 K2SO4 + 2 MnS04 + 8 H2O? 

Ans. 0.042236 g. 

412. A solution of potassium dichromate is equivalent to 
0.005000 g. of iron per cubic centimeter: 

K2Cr207+6 FeCl2+14 HC1 = 2 KC+2 CrCl3+6 FeCls+T H2O; 
if the solution is used to determine antimony: 
K2Cr207+3 SbCl3+14 HC1 = 3 SbCl5+2 CrCl3+2 KCl+7 H2O, 

what is the equivalent of antimony per cubic centimeter of this 
solution? Ans. 0.005381 g. 

413. A permanganate solution is equivalent to 0.005000 g. 
of iron per cubic centimeter according to the reaction: 

2 KMn04 + 10 FeS04 + 8 H2SO4 = K2SO4 + 2 MnS04 
-l-5Fe2(S04)3 + 8H20; 

if it is used to determine nitrous acid according to the equation: 

2 KMn04 + 5 HNO2 + 3 H2SO4 = K2SO4 + 2 MnS04 
+ 5HNO3 + 3H2O: 

(a) What is the value of the permanganate solution per cubic 
centimeter in terms of nitrous acid? {b) In terms of nitrous 
anhydride, (N2O3)? Ans. (a) 0.002105 g. 

(6) 0.001702 g. 

414. How many grams per cubic centimeter are contained in 
an N/2 solution of (a) potassium carbonate and (6) potassium 



VOLUMETRIC ANALYSIS 233 

bicarbonate, when the solutions are used to neutraUze acids? 
(The reaction results in decomposition to carbon dioxide.) 

Ans. (a) 0.03455 g. 
(6) 0.05054 g. 

415. How many grams per cubic centimeter are contained in 
a 0.1 N solution of acetic acid? Ans. p.006003 g. 

416. What is the normal quantity of (a) barium sulphate? 
(6) Barium chloride? (c) Silver chloride? 

Ans. (a) 116.72 g. 

(b) 104.15 g. 

(c) 143.34 g. 

417. How many grams per cubic centimeter in an N/10 solu- 
tion of (a) ferrous sulphate (FeS04 • 7 H2O), and (b) ferrous 
ammonium sulphate (FeS04 • (NH4)2S04 • 6 H2O), when used 
with an oxidizing agent? Ans. (a) 0.027802 g. 

(b) 0.039216 g. 

418. Potassium tetroxalate, KH3(C204)2 • 2 H2O, as an acid 
reacts : 

2 KH3(C204)2 + 6 NaOH = K2C2O4 + 3 Na2C204 + 6 H2O, 

and as a reducing agent: 

KH3(C204)2 + 2 Mn02 + 3 H2SO4 = 2 MnS04 + KHSO4 
+ 4CO2 + 4H2O. 

(a) How many grams of the crystallized salt are contained in a 
liter when used as a standard N/10 acid solution? (6) As a 
reducing agent? Ans. {a) 8.472 g. 

(6) 6.354 g. 

419. A solution of sodium hydroxide containing 0.02003 g. 
per cubic centimeter is to be used in titrating sulphuric acid. 
What weight of the acid should be taken so that the reading of 
the burette in cubic centimeters shall equal the percentage of 
sulphuric acid in the sample? Ans. 2.4553 g. 

420. A solution of potassium dichromate is equivalent to 
0.00559 g. of iron per cubic centimeter. How many grams of 



234 CHEMICAL CALCULATIONS 

iron ore must be taken so that each cubic centimeter used in the 
determination shall indicate (a) 0.1% Fe? (6) 0.5% Fe? 

Ans. (a) 5.59 g. 
(b) 1.118 g. 

421. When phenol-phthalein is used as an indicator, potas- 
sium dichromate may be titrated with an acid: 

H2SO4 + K2Cr207 = KHCrsOy + KHSO4. 
How many grams of potassium dichromate must be contained in 
a liter to make a normal solution? Ans. 147.10 g. 

422. Potassium ferrocyanide, K4Fe(CN)6 • 3 H2O, acts as a 
reducing agent : 

10 K4Fe(CN)6 + 2 KMn04 + 8 H2SO4 = 10 K3Fe(CN)6 
+ 6 K2SO4 + 2 MnS04 + 8 H2O. 
How much potassium ferrocyanide is contained in a cubic centi- 
meter of the normal solution? Ans. .42235 g. 

423. How much arsenic is shown by 12.00 cc. of N/20 iodine 
solution, the reaction being 

AS2O3 + 4 1 + 2 H2O = AS2O5 + 4 HI. 

Ans. 0.022488 g. 

424. To what weight of calcium oxide does 30.00 cc. of 0.1 N 
potassium permanganate correspond when the calcium has been 
separated as the oxalate, acidified and the oxalic acid titrated 
by potassium permanganate? Ans. 0.084105 g. 

425. (a) How many grams of silver nitrate in a cubic centi- 
meter of N/5 solution? (6) To how many grams of sodium 
chloride are 20.00 cc. of this~ solution equivalent? 

Ans. (a) 0.033978 g. 
(b) 0.2338 g. 

426. According to the reaction: 

NH4CNS + AgNOs = NH4NO3 + AgCNS. 

(a) What is the weight of ammonium sulpho cyanate in a cubic \ 
centimeter of a 0.1 N solution? (b) How many grams of silver j 
are shown by 15.00 cc. of 0.1 N ammonium sulpho cyanatf 
solution? Ans. (a) 0.007612 g. 

(6) 0.1618 g. 



VOLUMETRIC ANALYSIS . 235 

427. What is the value in terms of manganese dioxide, of a 
cubic centimeter of an N/10 solution of ferrous sulphate acting 
as a reducing agent, the reaction being 

2 FeS04 + Mn02 + 2 H2SO4 = MnS04 + Fe2(S04)3 + 2 H2O. 

Ans. 0.004347 g. 

428. Phosphoric acid may be determined by adding an excess 
of silver nitrate and sodium acetate, when the following reac- 
tion takes place: 

H3PO4 + 3 AgNOs + 3 CHsCOONa = Ag3P04 + 3 NaNOa 
+ 3 CH3COOH, 

the acetic acid being titrated with barium hydroxide, 

Ba(0H)2 + 2 CH3COOH = Ba(CH3COO)2 + 2 H2O. 

If 41.20 cc. of 0.1 N barium hydroxide are required to neutralize 
the acetic acid, how much phosphoric acid is present? 

Ans. 0.13467 g. 

429. Antimony may be determined by digestion with potas- 
sium iodide and hydrochloric acid, the iodine distilled over and 
collected in potassium iodide and titrated with sodium thiosul- 
phate. The reactions are 

KI + HCl = KCl + HI, 

4 HI + Sb205 = SbsOs + 2 H2O + 4 1, 

41 + 4 Na2S203 = 4 Nal + 2 Na2S406. 

If 30.25 cc. of thiosulphate are required to determine the iodine 
liberated, each cubic centimeter of which indicates 0.01282 g. 
of iodine, how much antimony is present? Ans. 0.18365 g. 

430. A solution of sulphuric acid contains 0.004912 g. of acid 
per cubic centimeter. What is the factor to 0.1 N? 

Ans. 1.0014. 

431. A solution of sulphuric acid is standardized against a 
standard solution of approximately 0.5 N sodium hydroxide, the 
factor of which is 0.9992 to 0.5 N. 45.00 cc. of the sodium 
hydroxide solution are required to neutrahze 44.70 cc. of the 
sulphuric acid solution. What is the factor of the sulphuric 
acid solution to 0.5 N? Ans. 1.006. 



236 CHEMICAL CALCULATIONS 

432. A solution of sodium hydroxide is to be standardized.' 
2.4530 g. of pure anhydrous sodium carbonate are weighed out 
and it requires 45.70 cc. of a solution of sulphuric acid for neu- 
tralization. 45.15 cc. of this solution of sulphuric acid require 
44.90 cc. of the solution of sodium hydroxide. What is the 
normality of this solution of sodium hydroxide? 

Ans. 1.018 N. 

433. A solution of potassium permanganate is found to be 
equivalent to 0.005600 g. of iron per cubic centimeter, (a) What 
is its normality? (b) If 20.20 cc. of this solution are used in a 
determination, to how many cubic centimeters of 0.1 N solution 
is this amount equivalent? Ans. (a) 0.1003 N. 

(6) 20.21 cc. 

434. 45.00 cc. of hydrochloric acid (factor to 0.1 N = 0.9987) 
require 87.25 cc. of a solution of barium hydroxide for neutral- 
ization, (a) What is the normality of the barium hydroxide? 
(6) What is the factor to 0.05 N? (c) How many grams of 
barium hydroxide are present per cubic centimeter? 

Ans. (a) 0.05151 N. 

(b) 1.030. 

(c) 0.004414 g. 

435. The factor of a solution of sodium chloride is 1.007 to 
0.1 N. 40.00 cc. of this solution are equivalent to 39.90 cc. 
of a solution of silver nitrate according to the reaction: 

AgNOg + NaCl = AgCl + NaNOs. 

(a) What is the factor of the solution of silver nitrate to 0.1 N? 

(b) How many grams per cubic centimeter does the silver ni- 
trate solution contain? Ans. (a) 1.0095. 

(6) 0.01715 g. • 

436. 0.2118 g. of arsenious oxide are weighed out to stand- 
ardize solutions of iodine and thiosulphate, the reactions being 

AS2O3 -f 4 1 + 2 H2O = AS2O5 + 4 HI, 
21 + 2 NasSoOa = 2 Nal + Na2S406. 

It requires 42.40 cc. of iodine to react with the arsenious oxide, 
(o) What is the normality of the iodine solution? (b) If 43.60 



i 



VOLUMETRIC ANALYSIS 237 

cc. of the thiosulphate solution are equivalent to 43.00 cc. of 
the iodine solution, what is the normality of the thiosulphate 
solution? Ans. {a) 0.1009 N. 

(6) 0.09961 N. 

437. A solution of sulphuric acid being precipitated with an 
excess of barium chloride gives 0.05841 g. of barium sulphate 
per cubic centimeter, (a) What is the normality of this solu- 
tion? (6) What is the factor to hah normal? (c) To how 
many cubic centimeters of half normal acid are 18.20 cc. of this 
solution equivalent? {d) If 2.5000 g. of sodium carbonate 
require 45.00 cc. of this solution of sulphuric acid for neutral- 
ization, what is the per cent of sodium carbonate in the sample? 

Ans. (a) 0.5004 N. 
(6) 1.001. 

(c) 18.22. 

(d) 47.73%. 

438. How many grams of sodium thiosulphate, Na2S203 • 
5 H2O must be present in a liter of solution to make it N/20, 
the reaction being 

2 Na2S203 + I2 = 2 Nal + Na2S406. 

Ans. 12.411 g. 

439. Oxalic acid may be used as an acid and as a reducing 
agent. When used as an acid, the reaction may be symboUzed: 

H2C2O4 + 2 NaOH = Na2C204 + 2 H2O. 

As a reducing agent, the reaction may be typified: 

H2C2O4 + = H2O + 2 CO2. 

How many grams of the crystallized oxalic acid, correspond- 
ing to the formula H2C2O4 • 2 H2O, must be taken per liter to 
prepare (a) a Hter of a 0.1 N solution in which the oxalic acid 
acts as an acid? (6) To prepare a 0.5 N solution, the acid 
acting as a reducing substance? Ans. (a) 6.3025 g. 

(6) 31.5125 g. 

440. Potassium tetroxalate, KHC2O4 • H2C2O4 • 2 H2O, crys- 
tallizes well and may be used to prepare standard solutions. 



238 CHEMICAL CALCULATIONS 

(a) How many grams of this salt should be taken to make a 
liter of normal solution, the salt acting as an acid according to 
the reaction: 

KHC2O4 . H2C2O4 + 3 NaOH = KNaC204 + Na2C204 + 3 H2O. 

(6) If used as a reducing solution, how many grams should be 
taken per liter to make a normal solution, the reaction being 
symbolized : 

KHC2O4 + 20 = KOH + H2O + 4 CO2. 

Ans. (a) 84.72 g. 
(6) 63.5325 g. 

441. Crude cream of tartar, KHC4H4O6, is to be titrated with 
0.4876 N NaOH solution. What weight of this substance 
must be taken for titration so that the burette reading shall 
indicate the percentage of KHC4H4O6 directly? 

KHC4H4O6 + NaOH = KNaC4H406 + H2O. 

Ans. 9.1737 g. 

442. A sample of oxalic acid, H2C2O4 • 2 H2O, is to be titrated 
with 0.1095 N potassium permanganate: 

2 KMn04 + 5 H2C2O4 + 3 H2SO4 = K2SO4 + 2 MnS04 
+ IOCO2 + 8H2O. 

What weight of the sample must be taken so that the burette 
reading multiplied by two shall give the percentage of C2H2O4 • 
2 H2O? Ans. 0.3451 g. 

443. Manganese is titrated with potassium permanganate in 
neutral solution according to the following reaction: 

2 KMn04 + 3 MnS04 + 2 H2O = K2SO4 + 5 MnOz + 5 H2SO4. 

(a) How many grams of potassium permanganate are contained 
in a normal solution according to this reaction? (6) How many 
grams of manganese are equivalent to a cubic centimeter of a 
tenth normal solution of potassium permanganate made up on 
this basis? (c) If the potassium permanganate is standardized 
for use in an acid solution, to how many grams of manganese 
per cubic centimeter is such a solution equivalent? (In acid 



VOLUMETRIC ANALYSIS 239 

solution, two molecules of potassium permanganate liberate 
five available oxygen atoms.) Ans. (a) 52.6767 g. 

(6) 0.0027465 g. 

(c) 0.001648 g. . 

444. A silver nitrate solution is 0.1032 N. What weight of 
sample of a chloride must be taken so that each cubic centimeter 
of silver nitrate solution used in titration shall indicate one per 
cent of chlorine? Ans. 0.3659 g. 

445. A sample containing iron is to be titrated with 0.2016 N 
potassium permanganate solution, (a) What weight of material 
should be weighed out for titration so that each cubic centi- 
meter used shall show one per cent of metallic iron? (6) So 
that the burette reading multiplied by two shall give the per- 
centage of iron as ferric oxide? Ans. (a) 1.1258 g. 

(b) 0.8048 g. 

446. How many cubic centimeters of a 1.093 N solution are 
equivalent to 42.82 cc. of a 1.047 N solution? Ans. 41.02 cc. 

447. How many cubic centimeters of a 0.02534 N solution 
are equivalent to 13.47 cc. of a 0.1073 N solution? 

Ans. 57.04 cc. 

448. How many cubic centimeters of a solution whose factor 
to N/2 is 1.003 are equivalent to 19.42 cc. of a solution whose 
factor to N/5 is 0.9942? Ans. 7.70 cc. 

449. How many cubic centimeters of a solution, the factor 
of which to N/10 is 0.9982, are equivalent to 15.94 cc. of a solu- 
tion whose factor to N/2 is 0.9743? Ans. 79.63 cc. 

450. (a) 30.00 cc. of a 0.5018 N solution of sulphuric acid 
neutralize 59.02 cc. of -a solution of sodium hydroxide. What 
is the normality of the sodium hydroxide solution? 

Ans. 0.2551 N. 

451. (b) 50.00 cc. of a solution of sulphuric acid yield 2.9250 
g. of barium sulphate. 30.00 cc. of this solution neutralize 
61.20 cc. of a solution of sodium hydroxide. What is the nor- 
mality of the sodium hydroxide solution? Ans. 0.2457 N. 



240 CHEMICAL CALCULATIONS 

452. Hydrogen peroxide is titrated with potassium perman- 
ganate according to the reaction: 

2 KMn04 + 3 H2SO4 + 5 H2O2 = K2SO4 + 2 MnS04 

+ 8 H2O + 5 O2. 

25,60 cc. of the hydrogen peroxide solution decolorized 45.90 cc. 
of the potassium permanganate solution, the strength of which 
is 0.6432 N. How many grams of hydrogen peroxide are present 
per liter? Ans. 1.9614 g. 

453. 3.0153 g. of a solution of hydrogen peroxide require 
53.15 cc. of a solution of 0.09920 N potassium permanganate to 
produce the pink color. The reaction is 

2 KMn04 + 3 H2SO4 + 5 H2O2 = K2SO4 + 2 MnS04 
+ 8H2O + 5O2. 

What is the percentage of hydrogen peroxide in this sample? 

Ans. 3.74%. 

454. Bleaching powder is to be analyzed for available chlo- 
rine. 7.0920 g. are rubbed up with water and made up to one 
liter. Of this suspension, 50.00 cc. are taken, treated with an 
excess of potassium iodide, and the liberated iodine titrated with 
N/10 solution of sodium thiosulphate. The reaction is 

Cl2 + 2KI = 2KCl + l2. 

If 35.50 cc. of the thiosulphate solution are required, what is 
the percentage of available chlorine? Ans. 35.50%. 

455. A sample of material containing pyrolusite (Mn02) 13 
treated with hydrochloric acid: 

Mn02 + 4 HCl = MnCl2 + CI2 + 2 H2O. 

The chlorine is led into a solution of potassium iodide liberal-, 
ing iodine: 

2KI + Cl2 = 2KCl + l2. 

The iodine is determined by a solution of sodium thiosulphate: 

I2 + 2 Na2S203 = 2 Nal + Na2S406. 

25.40 cc. of 0.1016 N thiosulphate solution are required. How 
much pyrolusite is present? Ans. 0.1122 g. 



VOLUMETRIC ANALYSIS 241 

456. (a) What is the equivalent in 0.5 N solution of 35.00 cc. 
of 0.1052 N solution? (6) How many cubic centimeters of 
0.2 N solution are equivalent to 42.00 cc. of a 0.5063 N solution? 

Ans. (a) 7.364 cc. 
(b) 106.3 cc. 

457. Manganese dioxide is reduced with an excess of oxalic 
acid (50.00 cc. of 0.2016 N) : 

Mn02 + H2SO4 + C2H2O4 = MnS04 + 2 CO2 + 2 H2O. 

The excess of oxalic acid requires 10.15 cc. of 0.2008 N potassium 
permanganate : 

2 KMn04 + 5 C2H2O4 + 3 H2SO4 = K2SO4 + 2 MnS04 
4-IOCO2 + 8H2O. 

How much manganese dioxide is present? Ans. 0.3495 g. 

458. 5.0000 g. of ammonium sulphate are treated in a flask 
with 100.00 cc. of sodium hydroxide solution (factor to 1 N = 
1.009). After boiling till all the ammonia is expelled, the solu- 
tion is titrated with sulphuric acid, 30.25 cc. (factor to 1 N = 
1.002) being required. What is the percentage of ammonia 
(NH3) indicated? Ans. 24.04%. 

459. One cubic centimeter of a potassium permanganate 
solution is equivalent to 0.005000 g. of iron. To 40.00 cc. of 
the permanganate solution an excess of potassium iodide is 
added which causes a liberation of iodine according to the reac- 
tion: 

2 KMn04 + 10 KI + 16 HCl = 12 KCl + 8 H2O + 2 MnCla -1- 10 1. 

The free iodine is titrated with a solution of sodium thiosulphate, 

2 Na2S203 + 2 1 = 2 Nal + Na2S406, 

35.90 cc. of the thiosulphate being required. What is the 
normality of the thiosulphate solution? Ans. 0.09977 N. 

460. 2.1200 g. of pure anhydrous sodium carbonate are 
weighed out and treated with 50.00 cc. of a solution of sulphuric 
acid: 

Na2C03 + H2SO4 = Na2S04 + H2O + CO2. 



242 CHEMICAL CALCULATIONS 

The excess of acid is titrated back with sodium hydroxide solu- 
tion, 10.05 cc. being required: 

H2SO4 + 2 NaOH = Na2S04 + 2 H2O. 

The solution of sulphuric acid and the sodium hydroxide solu- 
tion are titrated against each other when it is found that 45.00 
cc. of the sulphuric acid solution require 45.10 cc. of sodium 
hydroxide for neutralization. If the sodium hydroxide solution 
is used for determining oxalic acid (C2H2O4 • 2 H2O) , how many 
grams of oxalic acid are indicated by one cubic centimeter of the 
sodium hydroxide solution? 

H2C2O4 + 2 NaOH = Na2C204 + 2 H2O. 

Ans. 0.06286 g. 

461. Sulphur dioxide is determined by adding an excess of 
iodine (50.00 cc. of 0.1009 N): 

NasSOa + I2 + H2O = Na2S04 + 2 HI, 

the excess of iodine being titrated back with sodium thiosulphate 
(20.15 cc. of 0.1022 N): 

2 Na2S203 + I2 = Na2S406 + 2 Nal. 

How much sulphur dioxide is present? Ans. 0.09566 g. 

462. One gram of a sample of nitrate of soda is examined by 
treating with Devada's alloy (which produces nascent hydrogen 
when acted upon by sodium hydroxide) which reduces the 
nitrate to ammonia. The ammonia is collected in 100.00 cc. 
of 0.1157 N sulphuric acid. 10.15 cc. of 0.1063 N sodium hydrox- 
ide are necessary to neutralize the excess of sulphuric acid. What 
is the per cent of sodium nitrate in the sample? 

Ans. 89.18%. 

463. Potassium persulphate is estimated by potassium per- 
manganate indirectly. When treated with a ferrous salt in 
acid solution: 

K2S2O8 + 2 FeS04 = Fe2(S04)3 + K2SO4, 



VOLUMETRIC ANALYSIS 243 

the excess of ferrous sulphate being determined by perman- 
ganate : 

2 KMn04 + 10 FeS04 + 8 H2SO4 = K2SO4 + 2 MnS04 
+ 5 Fe2(S04)3 + 8 H2O. 

30.00 cc. of the ferrous sulphate solution are equivalent to 
31.25 cc. of 0.1000 N KMn04 solution. 30.00 cc. of the ferrous 
sulphate solution together with the sample of potassium per- 
sulphate require 7.35 cc. of 0.1000 N potassium permanganate 
solution. How much potassium persulphate is present? 

Ans. 0.3230 g. 

464. Iodine in iodides is determined by adding an excess of 
silver nitrate: 

KI + AgNOs = Agl + KNO3, 

and the excess of silver nitrate is found with potassium thio- 
cyanate : 

AgNOs + KCNS = KNO3 + AgCNS. 

42.00 cc. of silver nitrate solution are equivalent to 40.15 cc. 
of thiocyanate solution. 1.00 cc. of the silver nitrate solution 
contains 0.010820 g. of silver. How much potassium iodide is 
indicated if 32.00 cc. of the silver nitrate solution were added 
and 2.80 cc. of the thiocyanate solution were required for the 
back titration? Ans. 0.4841 g. 

465. Iron in the ferric condition is treated with an excess of 
sodium thiosulphate (50.00 cc.) : 

2 Na2S203 + 2 FeCla + 2 HCl = 2 FeCl2 + 4 NaCl + H2S4O6. 

The excess of thiosulphate is found to require 7.85 cc. of iodine 
solution. The solution of iodine was standardized by titrating 
it against a standard arsenic solution, each cubic centimeter of 
which contained 0.005162 g. of arsenious oxide: 

AS2O3 + 2 12 + 2 H2O = AS2O5 + 4 HI, 

45.00 cc. of the arsenic solution requiring 45.20 cc, of the iodine 
solution. The thiosulphate solution was standardized by 
titrating it against the iodine solution: 

I2 + 2 Na2S203 = 2 Nal + Na2S406, 



244 CHEMICAL CALCULATIONS 

45.00 cc. of the iodine solution requiring 45.95 cc. of the thio- 
sulphate solution. What is the weight of iron present? 

Ans. 0.2385 g. 

466. A solution of sodium carbonate and bicarbonate is to be 
examined for the content of these salts. Phenol-phthalein is 
added and the solution titrated with 0.1000 N hydrochloric acid 
in the cold, requiring 30.00 cc. Phenol-phthalein reacts neutral 
when all the carbonate is converted to bicarbonate : 

Na2C03 + HCl = NaHCOa + NaCl. 

Methyl orange is added which reacts acid when all the bicarbon- 
ate is decomposed: 

NaHCOs + HCl = NaCl + CO2 + H2O, 

35.00 cc. more being required. What are the amounts of car- 
bonate and bicarbonate of soda present? 

Ans. 0.3180 g. NazCOs, 
0.0420 g. NaHCOs. 

467. A sample of quicklime is to be analyzed for calcium 
oxide and calcium carbonate. 14.0000 g. are weighed out, 
slaked with warm water (free from carbon dioxide) and made 
up to 500 cc. Of this emulsion, 50 cc. are pipetted out and again 
made up to 500 cc, 50.00 cc. of this latter solution being taken 
for analysis. 60.00 cc, of N/10 hydrochloric acid solution are 
added and the whole heated to boiling or till there is no further 
evolution of carbon dioxide. 

Ca(0H)2 + 2 HCl = CaCl2 + 2 H2O, 
CaCOs + 2 HCl = CaCl2 + H2O + CO2. 

The excess of acid is found by titrating back with N/10 sodium 
hydroxide solution, 20 cc. being required, methyl orange being 
used as an indicator. A second portion of 50 cc. is taken and 
titrated with N/10 hydrochloric acid solution, using phenol- 
phthalein as an indicator, 35 cc. being required. Under these 
conditions only the calcium hydroxide is titrated. Calculate 
the percentages of calcium oxide and calcium carbonate in the 
quicklime. Ans. 70.09% CaO, 

17.87% CaCOs. 



VOLUMETRIC ANALYSIS 245 

468. 1.5000 g. of iron are dissolved in hydrochloric acid and 
treated with 0.7300 g. of a nitrate. 

2 MNO3 + 6 FeCl2 + 8 HCl = 2 MCI + 2 NO + 6 FeCU + 4 H2O. 

After reaction, the excess of ferrous chloride is determined by 
titration mth N/2 potassium permanganate. 

2 KMn04 + 10 FeCl2 + 16 HCl = 2 KCH- 2 MnCl2 
+ 10FeCl3 + 8H2O. 

An independent determination shows that 1.5000 g. of iron re- 
quire 53.55 cc. of the N/2 potassium permanganate solution. 
The excess of ferrous chloride requires 3.55 cc. of the perman- 
ganate solution, (a) If the nitrate taken is sodium nitrate 
(NaNOs), calculate the percentage present, (b) The per- 
centage of N2O5. Ans. (a) 97.05% NaNOs, 

(6) 61.66% N2O5. 

469. Phenol-phthalein is neutral to acid sodium carbonate. 
By using methyl orange, the acid sodium carbonate may be 
titrated to completion, i.e., to sodium chloride, carbon dioxide 
and water. A sample of sodium carbonate containing sodium 
hydroxide is titrated with hydrochloric acid (factor to 0.1 N = 
0.9985) using phenol-phthalein, when 50.00 cc. are required. 
Methyl orange is added and 20.00 cc. more hydrochloric acid 
are needed to produce acid reaction with this indicator. What 
are the amounts of sodium hydroxide and sodium carbonate 
present? Ans. 0.1198 g. NaOH, 

0.2117 g. NasCOs. 

470. Calculate the composition of a mixed acid from the fol- 
lowing : 

For total acid : 

Weight of acid taken 4. 8970 g. 

NaOH to neutralize 81 . 60 cc. 

For H2SO4 (other acids being driven off by evaporation) : 

Weight of acid taken 8.5420 g. 

NaOH to neutralize 80 . 75 cc. 



246 CHEMICAL CALCULATIONS 

For N2O3: 

Weight of acid taken 17 . OOO g. 

0.1 N KMn04 required . 10.00 cc. 

Factor NaOH solution to normal is 1.039. 

Ans. H2SO4 = 48.17%, 

HNO3 = 47.03%, 

N2O3 = 0.11%, 

H2O = 4.69%, 

100.00%. 

471. Calculate the composition of a mixed acid from the 
following : 

For total acid : 

Weight of acid taken 5 . 92I6 g. 

1 .012 N NaOH to neutrahze 99. 92 cc. 

For H2SO4 (other acids being driven off by evaporation) : 

Weight of acid taken 10 . 4387 g. 

1 .012 N NaOH to neutralize 104.00 cc. 

For N2O3: 

Weight of acid taken 18 . 000 g. 

. 1 N KMn04 required 9 . 00 cc. 

Ans. H2SO4 = 49.45%, 
HNO3 = 43.92%, 
N2O3 = 0.10%, 
H2O = 6.53 %, 

100.00%. 

472. Let 

X = percentage of free acetic anhydride (C2H30)20, 
y = percentage of total acidity calculated to acetic anhydride. 
Derive the formula 

X = 6.6643 (?/- 84.995). 

473. One gram of a sample of acetic anhydride containing 
acetic acid was treated with 200 cc. of a solution of barium 
hydroxide. 

(C2H30)20 + H2O = 2 CH3COOH, 

Ba(0H)2 + 2 CH3COOH = Ba(C2H302)2+ 2 H2O. 



VOLUMETRIC ANALYSIS 247 

200.00 cc. of the barium hydroxide solution was found to be 
equivalent to 190.00 cc. of an N/10 solution of hydrochloric acid. 
The excess of barium hydroxide was titrated back with N/10 
hydrochloric acid, 1.37 cc. being required. What are the per- 
centages of acetic anhydride and acetic acid present? ^ 

Ans. 74.99% (C2H30)20, 
25.01% CH3COOH. 

474. 0.8585 g. of a sample of acetic anhydride containing 
acetic acid is treated with 200.00 cc. of a solution of barium 
hydroxide. This volume of barium hydroxide solution is equiv- 
alent to 170.00 cc. of N/10 hydrochloric acid. The excess of 
barium hydroxide is titrated back with N/10 hydrochloric acid, 
8.46 cc. being required. Calculate the percentages of acetic 
anhydride and acetic acid present in this sample. (See the two 
problems above.) Ans. 73.41% (C2H30)oO, 

26.59% CH3COOH. 

475. Sodium hydroxide and trisodium phosphate are to bo 
determined in the presence of each other. Phenol-phthalein 
reacts neutral to disodium phosphate; therefore, in titrating a 
mixture of these two salts with sulphuric acid and phenol- 
phthalein: 

2 X NaOH -\-2y Na3P04 + (x + ?/) H2SO4 = 2y Na2HP04 
+ (x + y) Na2S04 + 2 x H2O. 

45.00 cc. of 0.5000 N H2SO4 are required for this titration. 
Methyl orange is now added, which is alkaline toward disodium 
phosphate, but neutral to monosodium phosphate. The titra- 
tion is continued with 0.5000 N H2SO4: 

2 y Na2HP04 + y H2SO4 = 2y NaH2P04 + y Na2S04, 

35.00 cc. being required. What are the amounts of trisodium 
phosphate and sodium hydroxide? Ans. 0.2000 g. NaOH, 

2.8707 g. Na3P04. 

476. (For the conduct of the sodium salts of phosphoric acid 
with indicators, see problem 475.) What is the composition of 
a solution of mixed tri and disodium phosphate if the phenol- 

* See problem above; also note similar case of oleum, p. 191. 



248 CHEMICAL CALCULATIONS 

phthalein titration required 25.00 cc. of 0.5000 N H2SO4 and the 
titration with methyl orange requires 35.00 cc. 0.5000 N H2SO4 
in addition? Ans. 0.7103 g. Na2HP04, 

2.0505 g. Na3P04. 

477. Oleum often contains sulphur dioxide. When the 
sample is dissolved in water and titrated with standard sodium 
hydroxide solution, using phenol-phthalein, the sulphur dioxide 
combines with the water, forming sulphurous acid, which takes 
up sodium hydroxide: 

H2SO3 + 2 NaOH = Na2SG3 + 2 H2O. 
What is the analysis of an oleum from the following data: 
For total acid as SO3: 

Weight of oleum taken 3 . 0570 g. 

1 N NaOH required 74. 30 cc. 

For sulphur dioxide: 

Weight of oleum taken 7 . 0510 g. 

0.1 N I solution required 46 . 80 cc. 

The reaction for the sulphurous acid and iodine is 

H2SO3 4- 12 + H2O = H2SO4 + 2 HI. 

Ans. Free SO3 = 80.31%, 

H2SO4 - 17.56%, 

SO2 - 2.13 %, 

100.00%. 

478. What is the composition of an oleum titrated with 
phenol-phthalein as an indicator, giving the following: 

For total acid: 

Weight of acid taken 5.0000 g. 

1.112 N NaOH to neutralize 99.95 cc. 

For sulphur dioxide: 

Weight of acid taken 5.0000 g. 

0.1 N I solution required 39 . 00 cc. 

Ans. Free SO3 - 34.17%, 
H2SO4 = 63.33%, 
SO2 = 2.50%. 



VOLUMETRIC ANALYSIS 249 

479. A solution shows 0,049205 g. of sulphuric acid per cubic 
centimeter. How many cubic centimeters of water must be 
added to a kilogram of this solution to make it LOGO N? 

Ans. 3.20 cc. 

480. How much 0.2019 N sodium hydroxide and water must 
be taken to make 5 liters of 0.1000 N sodium hydroxide? ^ 

Ans. 2391.2 cc. NaOH, 
2608.8 cc. H2O. 

481. 50.00 cc. of a solution (factor to N/10 = 1.005) corre- 
spond to 48.90 cc. of another solution. How many cubic centi- 
meters of water per liter must be added to this second solution 
to make it N/10? Ans. 28.00 cc. 

482. How many grams each of a 0.5012 N solution must be 
mixed with a 0.1078 N solution to make one kilogram of a 0.2000 
N solution? Ans. 765.6 g. 0.1078 N, 

234.4 g. 0.5012 N. 

483. How many pounds of 80.00% acetic acid must be added 
to 92.60% acetic acid to make 600 lbs. of 90.00% acetic acid? 

Ans. 123.8 lbs. 80.00%, 
476.2 lbs. 92.60%. 

484. How many pounds of water and how many pounds of 
60.00% sulphuric acid must be mixed to prepare 400 lbs. of a 
34.20% sulphuric acid? Ans. 288 lbs. 60% H2SO4, 

172 lbs. water. 

485. How many pounds of water must be added to 800 
pounds of 73.00% sulphuric acid to make the whole 70.00% 
sulphuric acid? Ans. 34.29 lbs. water. 

486. How much water must be added to 1000 cc. of a 0.1128 N 
solution to make it 0.1000 N? (When the densities of the solu- 
tions mixed, and the density of the resultant solution are very 
nearly the same, as in this case, volumes may be substituted in 
the formulas without sensible error.) 

Ans. 128.00 cc. water. 

1 Consider the densities of the two liquids and the resultant to be 
the same in this, and in other similar problems, unless otherwise 
mentioned. 



250 CHEMICAL CALCULATIONS 

487. How many pounds of 62.18% sulphuric acid and of 
98.00% sulphuric acid must be taken to make 1000 lbs. of 
93.00% acid? Ans. 139.6 lbs. 62.18% H2SO4, 

860.4 lbs. 98.00% H2SO4. 

488. How many pounds of 62.18% sulphuric acid must be 
added to 1000 lbs. of 98.00% sulphuric acid to dilute the whole 
to 93.00% acid? Ans. 162.2 lbs. 62.18% H2SO4. 

489. How much 0.1012 N solution must be added to 1000 cc. 
of a 0.5009 N solution to make it 0.2000 N? 

Ans. 3045.5 cc. 0.1012 N. 

490. How many pounds of 80.00% acetic acid and 60.00% 
acetic acid should be mixed to make 500 lbs. of 65.00% acetic 
acid?i Ans. 125 lbs. 80% CH3COOH, 

375 lbs. 60% CH3COOH. 

491. How many cubic centimeters of 0.0957 N and 0.1120 N 
solution must be taken to make 1000 cc. of 0.1000 N solution? 
(Consider the 0.0957 N solution as being strengthened by the 
0.1120 N solution.) Ans. 263.8 cc. 0.1120 N, 

736.2 cc. 0.0957 N. 

492. How many pounds of 80.00% sulphuric acid must be 
added to 980 lbs. of 35.00% sulphuric acid to strengthen the 
whole to 40.00% acid? Ans. 122.5 lbs. 80% H2SO4. 

493. How many pounds of a 20.00% hydrochloric acid must 
be added to 800 lbs. of a 43.00% hydrochloric acid to convert 
this weight into a 30.00% hydrochloric acid? 

Ans. 1040 lbs. 

494. How many pounds each of a 30.00% oleum and a 98.00% 
sulphuric acid must be mixed to prepare 100 lbs. of 100.00% 
sulphuric acid? (Calculate the percentages of SO3 present in 
each.) Ans. 22.82 lbs. 30.00% oleum, 

77.18 lbs. 98.00% H2SO4. 

^ In a problem of this kind in which a definite amount is called 
for, (a) of formula II of page 227 serves as well as (a) of for- 
mula III. In one case the formula gives the amount of diluting 
solution to add; in the other, the amount of strengthening solution 
required. 



VOLUMETRIC ANALYSIS 251 

495. A solution of potassium permanganate is equivalent 
to 0.01060 g. of Na2C204 per cubic centimeter. A weaker solu- 
tion is equivalent to 0.002120 g. of Na2C204 per cubic centi- 
meter. How many cubic centimeters of the stronger solution 
must be mixed with how many cubic centimeters of the weaker 
solution to prepare three liters of a solution equivalent to 
0.002792 g. of iron per cubic centimeter? 

Ans. 435 cc. stronger, 
2565 cc. weaker. 

496. A solution of potassium permanganate is equivalent to 
0.005600 g. of iron per cubic centimeter. What volume of this 
solution and what volume of water must be mixed to prepare 
three liters of a solution of which each cubic centimeter shall 
be equivalent to 0.005000 g. of iron? 

Ans. 2678 cc. sol., 
322 cc. H2O. 

497. A solution of sodium hydroxide is equivalent to 0.001000 g. 
of sulphiu-ic acid per cubic centimeter. How much water must 
be added to one liter of this solution to make it equivalent to 
0.001000 g. of nitric acid per cubic centimeter? 

Ans. 285.7 cc. 



CHAPTER X 

USE OF SPECIFIC GRAVITY TABLES AND 
ACID CALCULATIONS 

Calculation of Data in Specific Gravity Tables. — 

Large shipments of acid, particularly sulphuric acid, are 
usually billed and paid for on the basis of 66° Be., 60° Be., 
etc. It is, therefore, necessary to calculate the actual 
strength of the acid to its equivalent in 66° Be., or what- 
ever the strength basis of the acid is billed and paid for. 

The table of Ferguson and Talbot,^ for sulphuric acid 
will be used as a model, the same serving as an example 
of the use of such data. The methods of calculation of 
the data will be explained, the same being applicable to 
other solutions, the tables of which have not been so 
fully arranged. 2 

Specific gravity tables do not represent individual 
determinations for the densities listed,^ but are con- 
structed from curves and methods of interpolation al- 
ready considered.^ 

Twaddell's Hydrometers ^ consist of a series of hydrom- 
eters so arranged that the graduations permit of easy 
reading over a wide range. They are graduated only for 

1 See Appendix, also Chem. Ann., pp. 388-393. 

2 See Chem. Ann., pp. 392-456. 

3 The Baume is given in Ferguson and Talbot's table to two 
significant figures; it is accurate to two places of decimals. 

4 Pp. 32-41. 

^ See table of Ferguson and Talbot, Chem. Ann., pp. 388-392, 
column 3. 

252 



USE OF SPECIFIC GRAVITY TABLES 253 

liquids heavier than water. The unit of the Twaddell 
scale is 0.005 specific gravity; hence if X is the reading 
Twaddell, 

Sp.gr. = 1 + 0.005 X. 

The weight of a cubic foot of water at 60° F. is 62.37 
pounds. The weight of a cubic foot of a liquid is its 
specific gravity multiplied by 62.37.^ 

The acid content corresponding to 66° Be. (Oil of 
Vitriol, O. V.) has been carefully ascertained and found 
to be 93.19% H2SO4. A sample of sulphuric acid of 
65.75° Be. containing 91.80% H2S04^ is equivalent to 

|i^X 100 = 98.51% O.V.,« 

and as a cubic foot of liquid of 65.75° Be., which is equiva- 
lent to 1.8297 sp.gr., weighs 114.12 lbs. (62.37 X 1.8297),^ 
the number of pounds of Oil of Vitriol equivalent to one 
cubic foot of this acid is 

^1^X114.12 = 112.42 Ibs.^ 

The equivalent in 60° Be. acid (77.67% H2SO4) of an 
acid of 64° Be. (85.66%) H2SO4) is 

P|-5 X 100 = 110.29%,« 

1 See table of Ferguson and Talbot, Chem. Ann., pp. 388-392, 
column 3. 

2 Ibid., p. 392, column 4. 

3 lUd., p. 392, column 6. 
* Ihid., p. 392, column 5. 
^ Ibid., p. 392, column 7. 
6 Ibid., p. 393, column 3. 



254 CHEMICAL CALCULATIONS 

and as 60° Be. corresponds to 1.7059 sp. gr., the number 
of pounds of 60° Be. equivalent to a cubic foot of 64° Be. 
sulphuric acid is 

85.66 



77.67 



X 1.7059 X 62.37 = 123.14 Ibs.^ 



Use of Specific Gravity Tables. — As an example 
illustrating the use to which specific gravity tables may 
be put, suppose it is required to calculate the number of 
pounds of 50° Be. sulphuric acid in a shipment, the fol- 
lowing data being given: 

42 inches of sulphuric acid are drawn out of a tank at a 
temperature of 101° F. Suppose that it has been deter- 
mined by calculating the capacity of the tank from the 
inside measurements; that 1 inch in the depth of the 
tank corresponds to 50.00 cubic feet. A sample of 
the acid taken from the tank and brought into the labo- 
ratory showed 56.88° Be. at 92° F. Correction must be 
made for temperature in order to reduce to 60° F., the 
temperature for which the tables are constructed: 
92° - 60° = 32° difference.2 

From the table under the heading "Allowance for Temper- 
ature,"^ it is seen that the allowance for 60° Be. acid is 
0.026° Be. for each degree Fahrenheit, and that the correc- 
tion for 50° Be. acid is 0.028° Be. As the acid in ques- 
tion is about midway between these points, the allowance 
for each degree Fahrenheit is very nearly 0.027° B^. The 
correction for temperature is 

32 X 0.027 = 0.86° Be., 

^ See table of Ferguson and Talbot, Chem. Ann., p. 393, column 4; 
also from the product of the corresponding figure in column 5, p. 
392, and the fraction given. 

2 Ibid., p. 388, note to table. 

3 Ibid., p. 391. 



USE OF SPECIFIC GRAVITY TABLES 255 

and as the standard temperature, 60° F., is lower than 
92° F., the temperature at which the Baume of the sample 
was taken, the acid becoming denser as the temperature 
is lowered, this amount must be added. The Baume of 
the acid at 60° F. is, then, 

56.88 + 0.86 = 57.74° Be. at 60° F. 

The Baume of the acid at 101° F., the temperature at 
which the acid was drawn off, is calculated : 

101 - 60 - 41° F. difference, 
41 X 0.027 = 1.11° Be. correction; 

and as the density of the acid is lowered as the temper- 
ature is raised, 

57.74 - 1.11 = 56.63 Be. at 101° F. 

The easiest way to get the specific gravity corresponding 
to this degree Baume is by interpolating the given data : 

57° Be. = 1.6477 sp. gr.i 

56° Be. = 1.6292 sp. gr. 
diff. = 0.0185 sp. gr., 
56.63 - 56.00 = 0.63° Be. diff., 
0.0185 X 0.63 = 0.0117 sp. gr., 

1.6292 + 0.0117 = 1.6409 sp. gr. corresponding to 56.63° Be. 
Then, as 42 inches were drawn from the tank, the pounds 
drawn off are 

42 X 50.00 X 62.37 X 1.6409 = 214,920 lbs. 

As the acid is sold on the basis of 50° Be., the pounds of 
50° Be. corresponding to 57.74° Be. acid are easily found 
by interpolating from the table: 

1 See table of Ferguson and Talbot, Chem. Ann., p. 390, 
column 2. 



256 CHEMICAL CALCULATIONS 

58° Be. = 119.59% 50° Be. acid ^ 
57° Be. = 117.00 % 50° Be. acid 

diff . = 2.59%o 50° Be. acid, 
2.59 X 0.74 = 1.92, 
117 + 1.92 = 118.92% 50° Be. acid corresponding to 

57.74° Be. acid, 
214,920 X 1.1892 = 255,827 lbs. 50° Be. acid. 

Calculation of Mixed Acid. — '' Mixed Acid " is a 
commercial term, generally meaning a mixture of nitric 
and sulphuric acids, and is extensively used in manu- 
facturing processes. On account of the relative costs of 
concentrated and dilute nitric acids, the water called for 
is added in the form of dilute nitric acid, using a minimum 
of concentrated nitric acid and a maximum of dilute nitric 
acid. Water, as such, is seldom added. For example, 
let it be required to calculate the quantities of acids 
necessary to make a mixture ("mix") of 60,000 lbs. of a 
mixed acid to consist of 

Per cent 

H2SO4 (add as 98% H2SO4) 46.00 

HNO3 (add as 6L4% and as 95.5% HNO3) 49.00 

H2O 5.00 

100.00 

60,000 X 0.46 = 27,600 lbs. H2SO4 called for 
60,000 X 0.49 = 29,400 lbs. HNO3 called for 
60,000 X 0.05 = 3,000 lbs. H2O called for 
60,000 lbs. 

^^ = 28,163 lbs. 98% H2SO4 to take, 

60,000 - 28,163 = 31,837 lbs. still to add. 

1 See table of Ferguson and Talbot, Chem. Ann., p. 391, 
column 5. 



USE OF SPECIFIC GRAVITY TABLES 257 

Let X = number of pounds of 95.5% HNO3 to add; then 
31,837 — X will represent the number of pounds of 61.4% 
HNO3 to add. By algebra: 

0.955 a^ + 0.614 (31,837 - x) = 29,400. 
Solving: 

X = 28,891 lbs. 95.5% HNO3 to take, 
31,837 - 28,891 = 2946 lbs. 61.4%^ HNO3 to take. 

So, to make the mix, take 

H2SO4 28,163 lbs. 98% 

HNO3 28,891 lbs. 95.5% 

HNO3 2,946 lbs. 61.4% 

60,000 lbs. 

This same result might have been reached by means of 
the formulas given to adjust the strengths of liquids.^ 
29,400 lbs. of 100% nitric acid are called for; the weight 
of material still to be added, after the 98% sulphuric acid 
is added, is 31,837 lbs. as before. This makes 

29 400 

^^ X 100 = 92.35% HNO3 to be added. 

oi,oo/ 

To make 31,837 lbs. of an acid of this concentration from 
95.5% and 61.4% nitric acid, using III (a) of page 227, 

95 50 - 61 4 ^ ^^'^^^ " ^^'^^^ ^^^' ^^-^"^^ ^^^' ^^ 

take, as before. 
31,837 - 28,896 = 2941 lbs. 61.4% HNO3 to take, as 

before. 

Strengthening a Mixed Acid by Means of Oleum. — 
An example involving the use of oleum will next be 
considered. Let it be required to make 61,320 lbs. of a 
mixed acid of the composition: 

1 Pp. 227-228, II (a), or III (a), according to whether the acid is 
regarded as being diluted or strengthened. 



258 



CHEMICAL CALCULATIONS 



Per cent 

HNO3 (add as 94.5% HNO3) 56.00 

H2SO4 (add as 98.56% H2SO4 and as 20.00% oleum, 

a minimum of which is to be taken) 41 .00 

H2O 3.00 

100.00 

The tank in which the acid is to be mixed already contains 
2604 lbs. of the remains of a previous mix of the com- 
position : 

Per cent 

HNO3 52.00 

H2SO4 42.50 

H2O 5.50 

100.00 

Consequently, 





H2SO4 


HNO3 


H2O 


Called for 


Lbs. 
25,141 

1,107 
24,034 


Lbs. 

34,339 

1,354 

32,985 


Lbs. 

1840 


In tank 

To be added . 


143 
1697 







If the attempt were made to calculate the weights of 
acid to add by the previous method, it would be seen 
that the method would not work, as too much water 
would be added with the sulphuric acid and, hence, a 
nitric acid stronger than 94.5% HNO3 would have to be 
used to complete the mix; hence, oleum will have to be 
employed. Thus: 



24,034 
0.9856 



= 24,385 lbs. 98.56% H2SO4, 



24,385 - 24,034 = 351 lbs. H2O added with the 

98.56% H2SO4, 
1697 - 351 =. 1346 lbs. H2O remaining. 



USE OF SPECIFIC GRAVITY TABLES 259 

Adding this water with the nitric acid would call for a 
stronger nitric acid than the 94.5% HNO3, as is seen from 
the following: 

32,985 + 1346 = 34,331 lbs. HNO3 and H2O still to add, 

^^^ X 100 = 96.08% HNO3 required to complete 
the mix. 

Going back to the original figures after this preliminary 
calculation which has shown the necessity of using oleum; 
first calculating the weight of nitric acid to be added: 

^f^ = 34,905 lbs. 94.5% HNO3 to add, 
0.945 

34,905-32,985 = 1920 lbs. H2O added with the 94. 5%HN03. 

But the mix only calls for 1697 lbs. of water, hence 1920 — 
1697 == 223 lbs. of water will be added in excess. This 
water must be taken up with oleum. Now, to the acid 
already in the tank the following quantities of acid must 
be added: 

H2SO4 24,034 lbs. 100% H2SO4 

HNO3 32,985 lbs. 100% HNO3 

H2O 1,697 lbs. 100% H2O 

Total 58,716 

In adding 34,905 lbs. of 94.5% HNO3 there remain only 
58,716 - 34,905 = 23,811 lbs. of acid (H2SO4) to add. 
To adjust the proportions and not add more acid than 
called for is done by adding oleum which takes up water 
from the nitric acid. The percentage strength of the 
sulphuric acid requisite is 

g^ X 100 = 100.93% H2SO4. 



260 CHEMICAL CALCULATIONS 

The percentage of SO3 in 100.93% H2SO4 is 

ar) on cyj 

S^rSxlO''-93 = 82.39% SO3; 

in 98.56% H2SO4 the percentage of SO3 is 

lS); = SS>< 98-56 = 80.45% SO3; 

in 20.00% oleum the percentage of SO3 is ^ 

Per cent SO3 = 0.8163 (100 - 20) + 20 = 85.31% SO3; 
then, to make 23,811 lbs. of a 100.93% H2SO4, from 20.00% 
oleum and 98.56% H2SO4, require i^ 

CO QQ _ Qfl A-^ 

85 31 - 80 45 ^ ^^'^^^ ^ ^^^^ ^^'' ^^^^ ''^^''^' 

23,811 - 9506 = 14,305 lbs. 98.56% H2SO4. 

So, to make the mix, add to the acid already in the tank: 

HNO3 34,905 lbs. 94.50% 

H2SO4 14,305 lbs. 98.56% 

Oleum 9,506 lbs. 20.00% 

It is frequently desired to prepare a "mix" from a 
mixed acid already on hand by adding to it the requisite 
amounts of sulphuric acid and nitric acid to bring it up to 
the desired concentration. Thus, it may be required to 
fortify a " spent " mixed acid, or it may be that after add- 
ing the calculated amounts of ingredients to make a batch 
of mixed acid that the mixed acid resulting does not ana- 
lyze up to specifications. It must then be adjusted by a 
further addition of the deficient constituent. Thus, sup- 
pose a mixed acid of the following composition is desired : 

Per cent 

H2SO4 46.00 

HNO3 49.00 

H2O 5.00 

1 See p. 191. 2 See p. 228. 



USE OF SPECIFIC GRAVITY TABLES 261 

and there is on hand a supply of mixed acid of the com- 
position : 

Per cent 

H2SO4 45. 10 

HNO3 48.00 

H2O 6.90 

How many pounds of 98% sulphuric acid and 96% nitric 
acid should be added to the mixed acid for each 1000 lbs. 
of the corrected acid produced? An algebraic method of 
calculation is as follows : 

Let X = weight of mixed acid to take, 

y = weight of 9S% H2SO4 to take, 
z = weight of 96% HNO3 to take. 

The total sulphuric acid in the corrected mix must be 

(0.451) X = weight of H2SO4 added with the mixed acid, 
(0.98 ) y = weight of H2SO4 added as 98% H2SO4. 

The total nitric acid in the corrected mix must be 

(0.48) X = weight of HNO3 added with the mixed acid, 
(0.96) z = weight of HNO3 added as 96% HNO3. 

The total water in the corrected mix must be 

(0.069) X = weight of H2O added with 

the mixed acid, 

(1.00 - 0.98) y = (0.02) y = weight of H2O added with 

the 98%o H2S4O, 

(1.00 - 0.96) z = (0.04) z = weight of H2O added with 

the 96%o HNO3 acid. 

In each 1000 lbs. of the corrected mix there must be 

1000 X 0.46 = 460 lbs. H2SO4, 
1000 X 0.49 = 490 lbs. HNO3, 
1000 X 0.05 = 50 lbs. H2O. 



262 CHEMICAL CALCULATIONS 

Hence, the following equations result : 

(1) (0.451) X + (0.98) y = 460. 

(2) (0.48) X + (0.96) z = 490. 

(3) (0.069) X + (0.02) y + (0.04) z = 50. 

Expressing (1) and (2) in terms of x by solving for y and 
z in these equations : 

460- (0.451) X _ 460 (0.451) x 



(4) y 

(5) z 



0.98 0.98 0.98 

= 469.388 - (0.460204) x, 
490 - (0.48) X 490 (0.48) x 



0.96 0.96 0.96 

= 510.417 - (0.500000) X. 

Substituting in (3) gives 

(0.069) X + 0.02 [469.388 - (0.460204) x] 

+ 0.04 [510.417 - (0.500000) x] = 50, 
(0.069) X + 9.38776 - (0.00920408) x 
+ 20.41668 - (0.02) x = 50, 
(0.03979512) x = 20.19556, 

20.19556 ^„_ .^ „ . , . , ^ 

"" ^ 0.03979512 = ^^^'^^ ^^^' ^'^^^ ^"'^ ^^ 
take per 1000 lbs. 

Substituting this value of x in (4) and (5) gives 

y = 469.388 - (0.460204) (507.49) = 235.84 lbs. 

98% H2SO4 to take, 
z = 510.417 - (0.500000) (507.49) = 256.67 lbs. 

96% HNO3 to take. 

The ratios of these values may be used either to pre- 
pare a definite amount of mixed acid or to correct a defi- 
nite amount of " spent " acid. Knowing the ratios per 
1000 lbs. the quantities requisite for any weight of acid 
are readily calculated. After making a calculation, it is 



USE OF SPECIFIC GRAVITY TABLES 



263 



often a good plan to check up the results. This insures 
the accuracy of the calculation. Thus in the above cal- 
culation : 





H2SO4 


HNO2 


H2O 


Mixed acid 

98%H2S04 

96% HNO3 


Lbs. 

228.88 
231.12 


Lbs. 

243.60 


Lbs. 

35.01 
4.72 


246.40 
490.00 


10.27 










460.00 


50.00 



PROBLEMS 

498. What is the Twaddell reading corresponding to (a) 
1.6117 sp. gr.? (6) To 66.00° Be.? Ans. (a) 122.2 Tw., 

(6) 167.1 Tw. 

499. What is the reading of 141.2 Tw. in (a) specific gravity 
and (6) in Baume? Ans. (a) 1.7060 sp. gr., 

(6) 60.00° B^. 

500. {a) What is the percentage 0. V. (Oil of Vitriol, 93.19% 
H2SO4) .equivalent to 62.18% H2SO4? (6) What is the per- 
centage of 50° Be. sulphuric acid (62.18% H2SO4) equivalent 
toO.V.? Ans. {a) 66.72%, 

(6) 149.87%. 

501. (a) What is the equivalent m Oil of Vitriol (93.19% 
H2SO4) of 600 lbs. of a sulphuric acid of 89.55% H2SO4? (6) In 
50° Be. sulphuric acid (62.18% H2SO4)? Ans. (a) 576.6 lbs., 

(6) 864.12 lbs. 

502. Find the percentage of 100% H2SO4 equivalent to (a) a 
20.00% oleum? (6) To a 30.00% oleum? Ans. (a) 104.5%, 

(6) 106.75%. 

503. What is the percentage of Oil of Vitriol equivalent to 
25.00% oleum? (6) What is the percentage of 98.00% sul- 
phuric acid equivalent to a 35.00% oleum? 

Ans. (a) 113.34%, 
(6) 110.08%. 



264 CHEMICAL CALCULATIONS 

504. Knowing that 60° Be. sulphuric acid contains 77.67% 
H2SO4 and that 50° Be. sulphuric acid contains 62.18% H2SO4, 
what is the number of pounds of 50° B^. sulphuric acid equiv- 
alent to a cubic foot of 60° Be. sulphuric acid? 

Ans. 132.91 lbs. 

505. 50° Be. sulphuric acid contains 62.18% H2SO4, and 
52° Be. sulphuric acid contains 65.13% H2SO4. (a) To how 
many pounds of 50° Be. sulphuric acid are 350 cubic feet of 
52° Be. sulphuric acid equivalent? (6) If 60° Be. sulphuric 
acid contains 77.67% H2SO4, to how many pounds of 60° Be. 
sulphuric acid are 530 cubic feet of 52° Be. acid equivalent? 

Ans. (a) 35,647.5 lbs., 
(6) 43,216.2 lbs. 

506. Calculate the equivalent weight in terms of 60° B6. 
sulphuric acid equivalent to 2310 cubic feet, measured at 102 F., 
a sample of which showed 59.66° Be. at 80° F. 

Ans. 243,150 lbs. 

507. Calculate the weight of 50° Be. sulphuric acid equivalent 
to a shipment of 2160.61 cubic feet, measured at 120° F., a 
sample of which showed 56.14° Be. at 80° F.i 

Ans. 252,410 lbs. 

508. How many pounds of 60° Be. sulphuric acid are equiv- 
alent to a shipment of 2507 cubic feet measured at 92° F., a 
sample of which showed 65.52° Be. at 77° F.? 

Ans. 282,614 lbs. 

509. It is required to make 37,000 lbs. of a mixed acid of the 
composition: 

Per cent 

H3SO4 41 .00 

HNO3 52.00 

H2O 7.00 

there remaining in the mixing tank 6720 lbs. of an acid from a 
former mix of the composition: 

1 In commercial transactions, calculations are often carried to a 
degree of accuracy unwarranted by the accuracy of the readings. 



USE OF SPECIFIC GRAVITY TABLES 265 

Per cent 

H2SO4 42.00 

HNO3 52.54 

H2O 5.46 

How many pounds of 98.00% sulphuric acid, and 94.70% and 
61.40% of nitric acid must be added to the acid already in the 
tank to make this mix? Ans. 12,599.6 lbs. 98.00% H2SO4, 

14,574.3 lbs. 94.70%o HNO3, 
3,106.1 lbs. 61.40%, HNO3. 

510. If nitric acid to the extent of one per cent is lost in 
mixing the acids by reason of heat generated or other reasons, 
calculate the quantities called for in Problem 509 so that after 
mixing the acid will have the desired composition. 

Ans. 12,600 lbs. 98.00%c) H2SO4, 

15,689 lbs. 94.70% HNO3, 

1,991 lbs. 61.40%o HNO3. 

511. Calculate the amounts of acid required to make 34,000 
lbs. of a mixed acid of the composition : 

Per cent 

H2SO4 65.90 

HNO3 18.10 

H2O 16.00 

there remaining in the tank 3780 lbs. of an acid from a former 
mix, the composition of which is 

Per cent 

H2SO4 42.00 

HNO3 52.00 

H2O 6.00 

it being also desired to use in making this mix, in order to work 
it off, 7000 lbs. of an acid of 

Per cent 

H2SO4 64.00 

HNO3 28.00 

H2O 8.00 

The sulphuric acid is to be added as 93.20% H2SO4, and the nitric 
acid as 52.30% HNO3; water, if any be required, is to be added 



266 CHEMICAL CALCULATIONS 

as such. How many pounds of these substances must be taken, 
in addition to the acid already in the tank and the 7000 lbs. of 
the acid which it is desired to work off? 

Ans. 17,531 lbs. 93.2% H2SO4, 
4,260 lbs. 52.3% HNO3, 
1,429 lbs. water. 

512. Calculate how much 93.20% H2SO4, the acid to be 
worked off, and water must be added in problem (511) that a 
maximum amount of the acid to be worked off may be used, 
adding no 52.30% HNO3. 

Ans. 14,957 lbs. acid to be worked off, 
12,067 lbs. 93.2% H2SO4, 
3,196 lbs. water. 

513. How many pounds of 98.00% sulphuric acid, and 96.00% 
and 61.4% nitric acid must be taken to make 60,000 lbs. of a 
mixed acid of the composition: 

Per cent 

H2SO4 46.00 

HNO3 48.00 

H2O 6.00 

Ans. 28,163 lbs. 98.0% H2SO4, 

26,711 lbs. 96.0% HNO3, 

5,126 lbs. 61.4% HNO3. 

514. It is required to make a mixed acid of the composition: 

Per cent 

H2SO4 46.00 

HNO3 49.00 

H2O 5.00 

from 96.00% and 61.4% nitric acid and 98.00% sulphuric acid. 
How many pounds of each must be taken to prepare 60,000 lbs.? 

Ans. 28,163 lbs. 98.0% H2SO4, 

28,474 lbs. 96.0% HNO3, 

3,363 lbs. 61.4% HNO3. 

515. How many pounds of 95.00% nitric acid and 30.00% 
oleum must be added for each 1000 lbs. of a mixed acid of the 
composition: 



USE OF SPECIFIC GRAVITY TABLES 267 

Per cent 

H2SO4 43.00 

HNO3 51.00 

H2O 6.00 

to convert into a mixed acid of the composition: 

Per cent 

H2SO4 42.00 

HNO3 53.00 

H2O 5.00 

Ans. 137.07 lbs. 95% HNO3, 
71.38 lbs. 30% oleum. 
516. It is required to make 61,320 lbs. of a mixed acid of 
the composition: 

Per cent 

HNO3 56 . 00 

H2SO4 41.00 

H2O 3.00 

there remaining in the tank, from a former mix, 2604 lbs. of an 
acid of the composition: 

Per cent 

HNO3 52.00 

H2SO4 42.50 

H2O 5.50 

How many pounds of a 20.00% oleum, 98.56% sulphuric acid 
and 94.50 nitric acid must be added to the acid abeady in the 
tank? Ans. 9,678 lbs. 20.00% oleum, 

14,133 lbs. 98.56% H2SO4, 
34,905 lbs. 94.50% HNO3. 



388 



APPENDIX 



XXXV. — SUL 

By W. C. Ferguson 





Specific 






Weight of 




Pounds 


Degrees 


Gravity 


Degrees 


Per cent 


I Cu. Ft. 


Per cent 


0. V. in 


Baume. 


6o° 


Twaddell. 


H2SO4. 


in Lbs. 
Av. 


O.V.* 


I Cubic 
Foot. 





1.0000 


0.0 


0.00 


62.37 


0.00 


0.00 


1 


1.0069 


1.4 


1.02 


62.80 


1.09 


0.68 


2 


1.0140 


2.8 


2.08 


63.24 


2.23 


1.41 


3 


1.0211 


4.2 


3.13 


63.69 


3.36 


2.14 


4 


1.0284 


5.7 


4.21 


64.14 


4.52 


2.90 


5 


1.0357 


7.1 


5.28 


64.60 


5.67 


3.66 


6 


1.0432 


8.6 


6.37 


65.06 


6.84 


4.45 


7 


1.0507 


10.1 


7.45 


65.53 


7.99 


5.24 


8 


1.0584 


11.7 


8.55 


66.01 


9.17 


6.06 


9 


1.0662 


13.2 


9.66 


66.50 


10.37 


6.89 


10 


1.0741 


14.8 


10.77 


66.99 


11.56 


7.74 


11 


1.0821 


16.4 


11.89 


67.49 


12.76 


8.61 


12 


1.0902 


18.0 


13.01 


68.00 


13.96 


9.49 


13 


1.0985 


19.7 


14.13 


68.51 


15.16 


10.39 


14 


1.1069 


21.4 


15.25 


69.04 


16,36 


11.30 


15 


1.1154 


23.1 


16.38 


69.57 


17.58 


12.23 


16 


1 . 1240 


24.8 


17.53 


70.10 


18.81 


13.19 


17 


1 . 1328 


26.6 


18.71 


70.65 


20.08 


14.18 


18 


1.1417 


28.3 


19.89 


71.21 


21.34 


15.20 


19 


1.1508 


30.2 


21.07 


71.78 


22.61 


16.23 


20 


1.1600 


32.0 


22.25 


72.35 


23.87 


17.27 


21 


1 . 1694 


33.9 


23.43 


72.94 


25.14 


18.34 


22 


1 . 1789 


35.8 


24.61 


73.53 


26.41 


19.42 


23 


1 . 1885 


37.7 


25.81 


74.13 


27.69 


20.53 


24 


1 . 1983 


39.7 


27.03 


74.74 


29.00 


21.68 



Sp. Gr. determinations were made at 60° F., compared with 
water at 60° F. 

From the Sp. Grs., the corresponding degrees Baume were cal- 
culated by the following formula: Baume = 145 — 145/Sp. Gr. 

Baume Hydrometers for use with this table must be graduated 
by the above formula, which formula should always be printed on 
the scale. 

* 66° Baume = Sp. Gr. 1.8354 = Oil of Vitriol (O. V.). 
1 cu. ft. water at 60° F. weighs 62.37 lbs. av. 
Atomic weights from F. W. Clarke's table of 1901. O = 16. 
H2SO4 = 100 per cent. 
%H2S04 %0.V. %60° 
O. V. = 93.19 = 100.00 = 119.98 
60° = 77.67 = 83.35 = 100.00 
50° = 62.18 = 66.72 = 80.06 

(268) 



APPENDIX 



389 



PHURIC ACID 
AND H. P. Talbot 





* Freez- 


APPROXIMATE BOILING 


POINTS 


Degrees 
Baume. 


ing 

(Melting) 




50° B, 295° F. 






Point. F. 




60° 
61° 


' 386" 
' 400° 









32.0 




62° 


' 415° 






1 


31.2 




63° 


' 432° 






2 


30.5 




64° 


' 451° 






3 


29.8 




65° 


' 485° 






4 


28.9 




66° '' 538° " 




5 
6 
7 
8 
9 
10 
11 


28.1 

27.2 
26.3 
2^ 1 


FIXED POINTS 




Specific 


Per Cent 


Specific 


Per Cent 


24.0 
22.8 
21.5 
20.0 
18.3 


Gravity. 


H,S04. 


Gravity. 


H2SO4. 


1.0000 


.00 


1.5281 


62.34 


1.0048 


.71 


1.5440 


63.79 


12 
13 


1.0347 


5.14 


1.5748 


66.51 


1.0649 


9.48 


1.6272 


71.00 


14 


16.6 


1.0992 


14.22 


1.6679 


74.46 


15 


14.7 


1.1353 


19.04 


1.7044 


77.54 


16 


12.6 


1.1736 


23.94 


1.7258 


79.40 


17 


10.2 


1.2105 


28.55 


1.7472 


81.32 


18 


7.7 


1.2513 


33.49 


1.7700 


83.47 


19 


4.8 


1.2951 


38.64 


1.7959 


86.36 


20 


+ 1.6 


1.3441 


44.15 


1.8117 


88.53 


21 


- 1.8 


1.3947 


49.52 


1.8194 


89.75 


22 


- 6.0 


1.4307 


53.17 


1.8275 


91.32 


23 


-11 


1.4667 


56.68 


1.8354 


93.19 


24 


-16 


1.4822 


58. H 















Acids stronger than 66° Be. should have their percentage com- 
positions determined by chemical analysis. 

* Calculated from Pickering's results, Jour, of Lon. Ch. Soc, 
vol. 57, p. 363. 

Authorities — W. C. Ferguson; H. P. Talbot. 
This table has been approved and adopted as a standard by the 
Manufacturing Chemists' Association of the United States. 

W. H. Bower, 
Henry Howard, 
Jas. L. Morgan, 
Arthur Wyman, 
a. g. rosengarten, 
Executive Committee. 
New York, June 23, 1904. 



(269) 



390 



APPENDIX 





Specific 






Weight of 




Pounds 


Degrees 


Gravity 


Degrees 


Per Cent 


I Cu. Ft. 


Per Cent 


0. V. in 


Baume. 


6o° 


Twaddell. 


H2SO4. 


in Lbs. 
Av. 


O.V. 


I Cubic 
Foot. 


25 


1.2083 


41.7 


28.28 


75.36 


30.34 


22.87 


26 


1.2185 


43.7 


29.53 


76.00 


31.69 


24.08 


27 


1.2288 


45.8 


30.79 


76.64 


33.04 


25.32 


28 


1.2393 


47.9 


32.05 


77.30 


34.39 


26.58 


29 


1.2500 


50.0 


33.33 


77.96 


35.76 


27.88 


30 


1.2609 


52.2 


34.63 


78.64 


37.16 


29.22 


31 


1.2719 


54.4 


35.93 


79.33 


38.55 


30.58 


32 


1.2832 


56.6 


37.26 


80.03 


39.98 


32.00 


33 


1.2946 


58.9 


38.58 


80.74 


41.40 


33.42 


34 


1.3063 


61.3 


39.92 


81.47 


42.83 


34.90 


35 


1.3182 


63.6 


41.27 


82.22 


44.28 


36.41 


36 


1.3303 


66.1 


42.63 


82.97 


45.74 


37.95 


37 


1.3426 


68.5 


43.99 


83.74 


47.20 


39.53 


38 


1.3551 


71.0 


45.35 


84.52 


48.66 


41.13 


39 


1.3679 


73.6 


46.72 


85.32 


50.13 


42.77 


40 


1.3810 


76.2 


48.10 


86.13 


51.61 


44.45 


41 


1.3942 


78.8 


49.47 


86.96 


53.08 


46.16 


42 


1.4078 


81.6 


50.87 


87.80 


54.58 


47.92 


43 


1.4216 


84.3 


52.26 


88.67 


56.07 


49.72 


44 


1.4356 


87.1 


53.66 


89.54 


57.58 


51.56 


45 


1.4500 


90.0 


55.07 


90.44 


59.09 


53.44 


46 


1.4646 


92.9 


56.48 


91.35 


60.60 


55.36 


47 


1.4796 


95.9 


57.90 


92.28 


62.13 


57.33 


48 


1.4948 


99.0 


59.32 


93.23 


•63.65 


59.34 


49 


1.5104 


102.1 


60.75 


94.20 


65.18 


61.40 


50 


1.5263 


105.3 


62.18 


95.20 


66.72 


63.52 


51 


1.5426 


108.5 


63.66 


96.21 


68.31 


65.72 


52 


1.5591 


111.8 


65.13 


97.24 


69.89 


67.96 


53 


1.5761 


115.2 


66.63 


98.30 


71.50 


70.28 


54 


1.5934 


118.7 


68.13 


99.38 


73.11 


72.66 


55 


1.6111 


122.2 


69.65 


100.48 


74.74 


75.10 


56 


1.6292 


125.8 


71.17 


101.61 


76.37 


77.60 


57 


1.6477 


129.5 


72.75 


102.77 


78.07 


80.23 


58 


1.6667 


133.3 


74.36 


103.95 


79.79 


82.95 


59 


1.6860 


137.2 


75.99 


105.16 


81.54 


85.75 



APPENDIX 



391 





* Freezing 








Degrees 


(Melting) 








Baume. 


Point. 

°F. 


ALLOWANCE 
At 10° Be. .029° B 


FOR TEMPERATURE 


25 


-23 


e. or .00023 Sp.Gr.= l°F. 


26 


-30 


" 20° 


' .036° 


" .00034 ' 


1 << 


27 


-39 


'^30° 


' .035° 


'' .00039 ' 


= 1° '< 


28 


-49 


-40° 


' .031° 


" .00041 ' 


10a 


29 


-61 


''50° 


' .028° 


" .00045 ' 


1 f< 






"60° 


' .026° 


" .00053 ' 


iOCt 


30 


—74 


"63° 


' .026° 


" .00057 ' 


lO(C 


31 


-82 


" 66° 


' .0235° 


" .00054 " = 1° " 


32 


-96 








33 


-97 








34 
35 


-91 

—81 










Pounds 




Pounds 




oo 

36 


-70 


Per Cent 


6o° Baume 


Per Cent 


50° Baume 




37 


-60 


6o° 
Baume. 


in 
I Cubic 


50° 
Baume. 


in 
I Cubic 




38 


—53 




Foot. 




Foot. 




39 
40 


-47 
-41 












61.93 


53.34 


77.36 


66.63 




41 


-35 


63.69 


55.39 


79.56 


69.19 




42 


-31 


65.50 


57.50 


81.81 


71.83 




43 


-27 


67.28 


59.66 


84.05 


74.53 




44 


-23 


69.09 


61.86 


86.30 


77.27 




45 


-20 


70.90 


64.12 


88.56 


80.10 




46 


-14 


72.72 


66.43 


90.83 


82.98 




47 


-15 


74.55 


68.79 


93.12 


85.93 




48 


-18 


76.37 


71.20 


95.40 


88.94 




49 


-22 


78.22 


73.68 


97.70 


92.03 




50 


-27 


80.06 


76.21 


100.00 


95.20 




51 


-33 


81.96 


78.85 


102.38 


98.50 




52 


-39 


83.86 


81.54 


104.74 


101.85 




53 


-49 


85.79 


84.33 


107.15 


105.33 




54 


-59 


87.72 


87.17 


109.57 


108.89 




55 






89.67 


90.10 


112.01 


112.55 




56 




S 


91.63 


93.11 . 


114.46 


116.30 




57 




o 


93.67 


96.26 


117.00 


120.24 




58 




u 


95.74 


99.52 


119.59 


124.31 




59 


— 7 


W 


97.84 


102.89 


122.21 


128.52 





(271) 



392 



APPENDIX 



Degrees 


Specific 
Gravity 


Degrees 


Per Cent 


Weight of 
I Cu. Ft. 


Per Cent 


Pounds 
0. V in 


Baume. 


6o° 


Twaddell. 


H2SO4. 


in Lbs. 

Av. 


O.V. 


I Cubic 
Foot. 


60 


1.7059 


141.2 


77.67 


106.40 


83.35 


88.68 


61 


1.7262 


145.2 


79.43 


107.66 


85.23 


91.76 


62 


1.7470 


149.4 


81.30 


108.96 


87.24 


95.06 


63 


1.7683 


153.7 


83.34 


110.29 


89.43 


98.63 


64 


1.7901 


158.0 


85.66 


111.65 


91.92 


102.63 


641 


1.7957 


159.1 


86.33 


112.00 


92.64 


103.75 


64^ 


1.8012 


160.2 


87.04 


112.34 


93.40 


104 93 


64f 


1.8068 


161.4 


87.81 


112.69 


94.23 


106 19 


65 


1.8125 


162.5 


88.65 


113.05 


95.13 


107 54 


65i 


1.8182 


163.6 


89.55 


113.40 


96.10 


108.97 


651 


1.8239 


164.8 


90.60 


113.76 


97.22 


110.60 


fj 


1.8297 


165.9 


91.80 


114.12 


98.51 


112.42 


66 


1.8354 


167.1 


93.19 


114.47 


100.00 


114.47 



XXXVI. — FUMING SULPHURIC ACID AT 20° 
Cl. Winkler 



Specific 
Grav 

ity. 



1.835 
1.840 
1.845 
1.850 
1.855 
1.860 
1.865 



870 
875 
880 
885 
890 
895 
900 



Total 
SO.. 



75.31 

77.38 
79.28 
80.01 
80.95 
81.84 
82.12 
82.41 
82.63 
82.81 
82.97 
83.13 
83.43 
83.48 



100 Parts Contain 



Free 
SO3.* 



H2SO4. 



1.54 
2.66 

4.28 
5.44 
6.42 
7.29 
8.16 
9.34 
10.07 



92.25 
94.79 
97.11 
98.01 
99.16 
98.46 
97.34 
95.76 
94.56 
93.58 
92.71 
91.94 
90.66 
89.93 



Acid 

of 

66° B. 



99 

90.69 

83.08 

80.10 

76.38 

72.81 

71.71 

70.53 

69.35 

68.92 

68.27 

67.55 

66.81 

66.24 



Specific 
Grav 
ity. 



1.905 
1.910 
1.915 
1.920 
1.925 



1.930 



1.935 
1.940 
1.945 
1.950 
1.955 
1.960 
1.965 
1.970 



Total 
SO3. 



83.57 
83.73 
84.08 
84.56 
85.06 
85.57 
86.23 
86.78 
87.13 
87.41 
87.65 
88.22 
88.92 
89.83 



100 Parts Contain 



Free 
S03.=f 



10.56 
11.43 
13.33 
15.95 

18.67 
21.34 
25.65 
28.03 
29.94 
31.46 
32.77 
35.87 
39.68 
44.64 



H2SO4. 



89.44 
88.57 
86.67 
84.05 
81.33 
78.66 
74.35 
71.97 
70.06 
63.54 
67.23 
64.13 
60.32 
55.36 



Acid 

of 

66° B. 



65.68 
65.25 
63.84 
62.10 
59.90 
57.86 
55.21 
53.00 
51.60 
50.48 
49.52 
47.23 
44.42 
40.78 



This column gives the amount of SO3 which may be distilled off. 

(272) 



APPENDIX 



393 



Degrees 
Baume. 



61 
62 
63 
64 
64i 

641 

65 

65i 

651 

651 



* Freezing 

(Melting) 

Point. 



+ 12, 
27, 
39 
46 
46 
43 
41 
37 
33 
24 
13 

-1 

-29 



Per Cent 

6o° 
Baume. 



100.00 
102.27 
104.67 
107.30 
110.29 
111.15 
112.06 
113.05 
114.14 
115.30 
116.65 
118.19 
119.98 



Pounds 

6o° Baume 

in Cubic 

Foot. 



106.40 
110.10 

114.05 
118.34 
123.14 
124.49 
125.89 
127.40 
129.03 
130.75 
132.70 
134.88 
137.34 



Per Cent 

50° 
Baume. 



124.91 
127.74 
130.75 
134.03 
137.76 
138.84 
139.98 
141.22 
142.57 
144.02 
145.71 
147.63 
149.87 



Pounds 

50° Baume 

in Cubic 

Foot. 



132.91 

137.52 
142.47 
147.82 
153.81 
155.50 
157.25 
159.14 
161.17 
163.32 
165.76 
168.48 
171.56 



XXXVII. — SULPHURIC ACID 

94-100% H2SO4 

By H. B. Bishop 

The acid used in this table was prepared from Baker and Adam- 
eon's c.p. sulphuric acid 95 per cent, which was strengthened to 
100 per cent by the addition of fuming sulphuric acid made by dis- 
tilling fuming acid (70 per cent free SO3) into a portion of the 95 per 
cent c.p. acid. The final acid was tested for impurities: residue 
upon evaporation, chlorine, niter and sulphur dioxide. The only 
impurity found was a trace of sulphur dioxide (0.001 per cent) 
which was less than the sensitiveness of the determination. 

The analytical and specific gravity determinations, and the allow- 
ance for temperature were made in the same manner, and with the 
same accuracy as in the sulphuric acid table adopted in 1904, the 
specific gravity 1.8354 and 93.19 per cent H2SO4 being' taken as a 
standard. 

The actual determinations were made within a few hundredths 
of a per cent of the points given in the table, the even percentages 
being calculated by interpolation. 



Per Cent H2SO4. 


Sp. Gr. at 
60° F. 


Allowance for Temperature. 


66° Be. 93.19 


1.8354 


At 94% 0.00054 sp. gr. = 1° F. 


94.00 


1.8381- 


At 96% 0.00053 sp. gr. = 1°F. 


95.00 


1.8407 


At 97.5% 0.00052 sp. gr. = 1° F. 


96.00 


1.8427 


At 100% 0.00052 sp. gr. = 1° F. 


97.00 


1.8437 




97.50 


1.8439 




98.00 


1.8437 




99.00 


1.8424 




100.00 


1.8391 





(273) 



384 



APPENDIX 





TENSION OF AQUEOUS VAPOR 




Temp. 


Press., 


Temp. 


Press., 


Temp. 


Press., 


°C. 


mm. 


°C. 


mm. 


°C. 


mm. 


0° 


4.60 


12° 


10.48 


24° 


22.18 


1° 


4.92 


13° 


11.19 


25° 


23.55 


2° 


5.29 


14° 


11.94 


26° 


24.99 


3° 


5.68 


15° 


12.73 


27° 


26.50 


4° 


6.09 


16° 


13.57 


28° 


28.10 


5° 


6.53 


17° 


14.45 


29° 


29.78 


6° 


7.00 


18° 


15.38 


30° 


31.56 


7° 


7.49 


19° 


16.37 


31° 


33.42 


8° 


8.02 


20° 


17.41 


32° 


35.37 


9° 


8.58 


21° 


18.50 


33° 


37.43 


10° 


9.18 


22° 


19.66 


34° 


39.59 


11° 


9.81 


23° 


20.88 


35° 


41.85 



Mensuration 

TT = 3.1416. 

Length of circle (radius r) = 2 irr. 

Area of triangle (base h, altitude a) = — • 

Area of circle (radius r) = ttt^. 

Surface of sphere (radius r) = 4 -wr^. 

Surface of cyKnder (radius r, height h) = 2 irr {r -[- h) . 

Surface of cone (radius r, slant height I) = wr {r + ^). 

4 
Volume of sphere (radius r) = - irr^. 

o 

Volume of cylinder (radius r, height h) = irr%. 



Gases 

(Weight at Standard Conditions, 0° C. and 760 mm.) 

Air 1.2926 grams 

Carbon dioxide 1 . 9768 grams 

Hydrogen , 0.089873 grams 

Nitrogen 1 . 2507 grams 

Oxygen 1 , 4292 grams 



Miscellaneous 

One pound avoirdupois = 453.59 grams. 

One meter = 39.37 inches. 

One gallon = 231 cubic inches. 

One cubic foot of water at 60° F. weighs 62.37 pounds. 



(274) 



INDEX 



Analysis: 

Direct gravimetric, 173. 

Elimination of constituent, 
176. 

Indirect gravimetric, 177. 

Volumetric, 202. 
Approximate numbers: 

Addition, 28. 

Definition, 26. 

Division, 30. 

Multiplication, 29. 

Rules, 31. 

Subtraction, 31. 
Atomic weights: 

Calculation, 145. 

Relation to valence, 142. 

Table, back of cover. 

Unit, 91, 141. 

Barometer : 

Correction, 84. 
Baume : 

Conversions, 65. 

Sulphuric acid, 35. 

Equivalent : 

Definition, 12. 

Values of solution, 204. 
Extrapolation : 

Method, 40. 

Factor: 

Chemical, 9. 
General, 15. 
Weights, 176, 205. 



Formula : 

From composition, 149. 
Of compounds, 148. 
Of minerals, 157. 

Gas: 

Analysis, 88. 
Moist, 85. 
Thermometer, 81. 
Volume from weight of sub- 
stance, 101. 

Heat: 

British Thermal Unit, 49. 

Calorie, 49. 

Of combustion, 51. 

Of fusion, 50. 

Of vaporization, 50. 

Specific, 50. 
Hydrometers : 

Baume, 64, 

Constant volume, 63. 

Constant weight, 62. 

Interpolation : 
Method, 37. 

Law: 

Avogadro, 90. 

Boyle, 80. 

Charles, 80. 

Charles and Boyle combined, 

82. 
Constancy of composition, 1. 
Constancy of mass, 1. 



275 



276 



INDEX 



Law (continued): 

Dalton, 107. 

Dulong and Petit, 143. 

Gay-Lussac, 87. 

Graham, 94. 
Liter: 

Mohr, 106. 

Standard, 104. 

Mixed acid: 

Analysis, 218. 

Calculations, 256. 

Strengthening of, 257. 
Molecular weights : 

From elevation of B.P., 146. 

From depression of F.P., 146. 

From vapor density, 91. 

Of an organic acid, 147. 

Of an organic base, 147. 

Normality: 

Calculation of, 213. 

Equivalent, 214. 

Factor, 213. 

Of acids and bases, 208. 

Of oxidizing substances, 209. 

Of salt solutions, 210. 

Oleum: 

Calculation of strength, 191. 
Volumetric titration of, 222. 

Ratios : 

Chemical, 2. 
General, 15. 

Solutions: 

Adjustment of strength, 225. 
Calculation of normality, 213. 
Calculation with normal, 210. 



Solutions (continued) : 

Equivalent values, 204. 

Methods of standardizing, 203. 

Normal, 207. 

Single factor, 202. 
Specific gravity: 

Calculation of tables, 252. 

Definition, 59. 

Influence of temperature, 66. 

Liquids, by pyknometer, 62. 

Liquids, by sinker, 62. 

Powders, 61. 

Solid, heavier than water, 60. 

Solid, lighter than water, 61. 

Use of tables, 254. 

Temperature: 

Absolute scale, 49. 

Centigrade scale, 48. 

Conversions, 51. 

Fahrenheit scale, 48. 

Reaumur scale, 49. 

Standard conditions, 83. 
Titration : 

Mixed acid, 218. 

Oleum, 222. 

With two indicators, 220. 

With two solutions, 216. 

Valence: 

Calculation of, 144. 

Definition, 141. 
Vapor density: 

Definition, 92. 

Measurement, 96. 

Weight: 

In vacuo, 102. 
Combining, 141. 



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worsted, cotton and silk materials). New Edition, re- 
zvritten and enlarged. 39 colored plates. 367 illustra- 
tions. 8vo. cloth. 369 pp. net, $6.00 

BECHHOLD, H. Colloids in Biology and Medicine. 
Translated by J. G. Bullowa, M.D. In Press. 

BEEKMAN, J. M. Principles of Chemical Calculations. 

In Press. 

BENNETT, HUGH G. The Manufacture of Leather, 
no illustrations. 8vo. cloth. 438 pp. net, $4.50 

BERNTHSEN, A. A Text-book of Organic Chemistry. 
English translation. Edited and revised by J. J. Sud- 
borough. Illus. i2mo. cloth. 690 pp. net, $2.50 

BERSCH, J. Manufacture of Mineral Lake Pigments. 
Translated by A. C. Wright. 43 illustrations. 8vo. 
cloth. 476 pp. net, $5,00 

BEVERIDGE, JAMES. Papermaker's Pocketbook. Spe- 
cially compiled for paper mill operatives, engineers, 
chemists and office officials. Second and Enlarged 
Edition. Illus. i2mo. cloth. 211 pp. net, $4.00 

BIRCHMORE, W. H. The Interpretation of Gas Analyses. 
Illustrated. i2mo. cloth. 75 pp. net, $1.25 

BLASDALE, W. C. Principles of Quantitative Analysis. 
An introductory course. 70 illus. 5^x7^. cloth. 
404 pp. net, $2.50 

BLt/CHER, H. Modern Industrial Chemistry. Trans- 
lated by J. P. Millington. Illus. 8vo. cloth. 795, 
pp. net, $7.50 

BL¥TH, A. W. Foods: Their Composition and Analysis. 
A manual for the use of analytical chemists, with an 
introductory essay on the History of Adulterations. 
Sixth Edition, thoroughly revised, enlarged and re- 
written. Illustrated. 8vo. cloth. 634 pp. $7.50 

Poisons : Their Effects and Detection. A manual for 

the use of analytical chemists and experts, with an 



LIST OF CHEMICAL BOOKS 



introductory essay on the Growth of Modern Toxicol- 
ogy. Fourth Edition, revised, enlarged and rewritten. 
Illustrated. 8vo. cloth. yy2 pp. $7.50 

BOCKMANN", F. Celluloid ; Its Raw Material, Manufac- 
ture, Properties and Uses. 49 illustrations. i2mo. cloth. 
120 pp. net, $2.50 

BOOTH, WILLIAM H. Water Softening and Treatment. 
91 illustrations. 8vo. cloth. 310 pp. net, $2.50 

BOURCART, E. Insecticides, Fungicides, and Weed 
Killer?. Translated by D. Grant. 8vo. cloth. 500 pp. 

net, $4.50 

BOTJRRY, EMILE. A Treatise on Ceramic Industries. 
A complete manual for pottery, tile, and brick manu- 
facturers. A revised translation from the French by 
Alfred B. Searle. 308 illustrations. 12 mo. cloth. 
488 pp. net, $5.00 

BRISLEE, F. J. An Introduction to the Study of Fuel. 
A text-book for those entering the engineering, chem- 
ical and technical industries. 60 ill. 8vo. cloth. 293 
pp. (Outlines of Industrial Chemistry.) net, $3.00 

BRUCE, EDWIN M. Detection of the Common Food 
Adulterants. Illus. i2mo. cloth. 90 pp. net, $1.25 

BUSKETT, E. W. Fire Assaying. A practical treatise on 
the fire assaying of gold, silver and lead, including 
descriptions of the appliances used. Illustrated. i2mo. 
cloth. 112 pp. net, $1.25 

BYERS, HORACE G., and KNIGHT, HENRY G. Notes 

on Qualitative Analysis. Second Edition, revised. 
8vo. cloth. 192 pp. net, $1.50 

CAVEN, R. M., and LANDER, G. D. Systematic Inor- 
ganic Chemistry from the Standpoint of the Periodic 
Law. A text-book for advanced students. Illustrated. 
i2mo. cloth. 390 pp. • net, $2.00 



4 D. VAN NOSTRAND COMPANY'S 

CHRISTIE, W. W. Boiler-waters, Scale, Corrosion, Foam- 
ing, yy illustrations. 8vo. cloth. 242 pp. net, $3.00 

Water, Its Purification and Use in the Industries. 

79 illus., 3 folding plates, 2 colored inserts. i2mo. 
cloth. 230 pp. net, $2.00 

CHURCH'S Laboratory Guide. A manual of practical 
chemistry for colleges and schools, specially arranged 
for agricultural students. Ninth Edition^ revised and 
partly rewritten by Edward Kinch. Illustrated. 8vo. 
cloth, 365 pp. net, $2.50 

CORNWALL, H. B. Manual of Blow-pipe Analysis. 
Qualitative and quantitative. With a complete system 
of determinative mineralogy. Sixth Edition, revised. 
70 illustrations. 8vo. cloth. 310 pp. net, $2.50 

CROSS, C. F., BEVAN, E. J., and SINDALL, R. W. 
Wood Pulp and Its Uses. With the collaboration of 
W. N. Bacon. 30 illustrations. i2mo. cloth. 281 
pp. (Van Nostrand's Westminster Series.) net, $2.00 

d'ALBE, E. E. F. Contemporary Chemistry. A survey 
of the present state, methods, and tendencies of chemi- 
cal science. i2mo. cloth. 172 pp. net, $1.25 

DANBY, ARTHUR. Natural Rock Asphalts and Bitu- 
mens. Their Geology, History, Properties and Indus- 
trial Application. Illustrated. i2mo. cloth. 254 pp. 

net, $2.50 

DEERR, N. Cane Sugar. 280 illustrations. 6>^x9^. 
•cloth. 608 pp. net, $7.00 

DUMESNY, P., and NOYER, J. Wood Products, Dis- 
tillates and Extracts. Translated by D. Grant. 103 
illustrations. 8vo. cloth. 320 pp. net, $4.50 

DUNSTAN, A. E., and THOLE, F. B. A Text-book of 
Practical Chemistry for Technical Institutes. 52 illus- 
trations. i2mo. cloth. 345 pp. net, $1.40 

DYSON, S. S., and CLARKSON, S. S. Chemical Works, 
Their Design, Erection, and Equipment. 80 illustra- 
tions, 9 folding plates. 8vo. cloth. 220 pp. net, $7.50 



LIST OF CHEMICAL BOOKS 



ELIOT, C. W., and STORER, F. H. A Compendious Man- 
ual of dualitative Chemical Analysis. Revised with 
the co-operation of the authors, by William R. 
Nichols. Twenty-second Edition, newly revised by 
W. B. Lindsay. 111. i2mo. cloth. 205 pp. net, $1.25 

ELLIS, C. Hydrogenation of Oils, Catalysis and Catalyzers, 
and the Generation of Hydrogen. 145 ill. 6x9. cloth. 
350 pp. net. $4.00 

ENNIS, WILLIAM D. Linseed Oil and Other Seed Oils. 
An industrial manual. 88 illustrations. 8vo. cloth. 
336 pp. net, $4.00 

ERMEN, W. F. A. The Materials Used in Sizing. Their 
chemical and physical properties, and simple methods 
for their technical analysis and valuation. Illustrated. 
i2mo. cloth. 130 pp. net, $2.00 

FAY, IRVING W. The Chemistry of the Coal-tar Dyes. 
8vo. cloth. 473 pp. net, $4.00 

FERNBACH, R. L. Chemical Aspects of Silk Manu- 
facture. i2mo. cloth. 84 pp. net, $1.00 

^Glue and Gelatine. A practical treatise on the 

methods of testing and use. Illustrated. 8vo. cloth. 
208 pp. net, $3.00 

FISCHER, E. Introduction to the Preparation of Or- 
ganic Compounds. Translated from the new (eighth) 
Germ.an edition by R. V. Stanford. Illustrated. 
i2mo. cloth. 194 pp. net, $1.25 

FOYE, J. C. Chemical Problems. Fourth Edition, revised 
and enlarged. i6mo. cloth. 145 pp. (Van Nos- 
trand Science Series, No. 69.) $0.50 

FRANZEN, H. Exercises in Gas Analysis. Translated 
from the first German edition, with corrections and 
additions by the author, by Thomas Callan. 30 dia- 
grams. 5x7^. cloth. 127 pp. net, $1.00 

FRITSCH, J. The Manufacture of Chemical Manures. 
Translated from the French, with numerous notes, by 



6 D. VAN NO STRAND COMPANY'S 

Donald Grant. 69 illiis., 108 tables. 8vo. cloth. 
355 pp. net, $4.00 

GEOSSMANN, J. Ammonia and Its Compounds. Illus- 
trated. i2mo. cloth. 151 pp. net, $1.25 

HALE, WILLIAM J. Calculations in General Chemistry. 
With definitions, explanations and problems. Second 
Edition, revised. i2mo. cloth. 185 pp. net, $1.00 

HALL, CLARE H. Chemistry of Paints and Paint Ve- 
hicles. 8vo. cloth. 141 pp. net, $2.00 

HILDITCH, T. P. A Concise History of Chemistry. 
16 diagrams. i2mo. cloth. 273 pp. net, $1.25 

HOPKINS, N. M. Experimental Electrochemistry : Theo- 
retically and Practically Treated. New Edition. 

In Press. 

HOULLEVIGUE, L. The Evolution of the Sciences. 
8vo. cloth. 377 pp. net, $2.00 

HtJBNER, JULIUS. Bleaching and Dyeing of Vegetable 
Fibrous Materials. 95 illus. (many in two colors). 
8vo. cloth. 457 pp. net, $5.00 

HUDSON, 0. F. Iron and Steel. An introductory text- 
book for engineers and metallurgists. With a section 
on Corrosion by Guy D. Bengough. 47 illus. 8vo. 
cloth. 184 pp. net, $2.00 

HURST, GEO. H. Lubricating Oils, Fats and Greases. 
Their origin, preparation, properties, uses, and analy- 
sis. Third Edition, revised and enlarged, by Henry 
Leask. 74 illus. 8vo. cloth. 405 p. net, $4.00. 

HURST, G. H., and SIMMONS, W. H. Textile Soaps and 
Oils. Second Edition, revised and partly rezvritten. 
II illustrations. 5^x8%. 204 pp. net, $3.00 

HYDE, FREDERIC S. Solvents, Oils, Gums, Waxes and 
Allied Substances. 5^x8>^. cloth. 182 pp. 

net, $2.00 

INGLE, HERBERT. Manual of Agriculturd Chemistry. 
Illustrated. 8vo. cloth. 388 pp. -aet; $3.00 



LIST OF CHEMICAL BOOKS 



JOHNSTON, J. F. W. Elements of Agricultural Chem- 
istry. Revised and lewritten by Charles A. Cameron 
and C. M. Aikman. Nineteenth Edition. Illustrated. 
i2mo. cloth. 502 pp. $2.60 

JONES, HARRY C. A New Era in Chemistry. Some of 
the more important developments in general chemis- 
try during the last quarter of a century. Illustrated. 
i2mo. cloth. 336 pp. net, $2.00 

KEMBLE, W. F., and UNDERHILL, C. R. The Periodic 
Law and the Hydrogen Spectrum. Illustrated. 8vo. 
paper. 16 pp. net, $0.50 

KERSHAW, J. B. C. Fuel, Water, and Gas Analysis, for 
Steam Users. 50 ill. 8vo. cloth. 178 pp. net, $2.50 

Electro-Thermal Methods of Iron and Steel Produc- 
tion. With an introduction by Dr. J. A. Fleming, 
F.R.S. 50 tables, 92 illustrations. S/^^^/i- cloth. 
262 pp. net, $3.00 

KNOX, JOSEPH. Physico-chemical Calculations. i2mo. 
cloth. 196 pp. net, $1.00 

The Fixation of Atmospheric Nitrogen. Illustrated. 

5x7^. cloth. 120 pp. (Van Nostrand's Chemical 
Monographs.) net, $0.75 

KOLLER, T. Cosmetics. A handbook of the manufac- 
ture, employment and testing of all cosmetic materials 
and cosmetic specialties. Translated from the German 
by Charles Salter. 8vo. cloth. 262 pp. net, $2.50 

Utilization of Waste Products. A treatise on the 

rational utilization, recovery and treatment of waste 
products of all kinds. Second Revised and Enlarged 
Edition. 22 illustrations. 5%x8%. cloth. 336 pp. 

net, $3.00 

KREMANN, R. The Application of Physico-chemical 
Theory to Technical Processes and Manufacturing 
Methods. Authorized translation by Harold E. Potts, 
M.Sc. 35 diagrams. 8vo. cloth. 215 pp. net, $2.50 



8 D. VAN NOSTRAND COMPANY'S 

KRETSCHMAR, KARL. Yarn and Warp Sizing in All 
Its Branches. Translated from the German by C. 
Salter. 122 illus. 8vo. cloth. 192 pp. net, $4.00 

LAMBORN, L. I. Modern Soaps, Candles and Glycerin. 
224 illustrations. 8vo. cloth. 700 pp. net, $7.50 

Cotton Seed Products. 79 illus. 8vo. cloth. 253 pp. 

net, $3.00 

LASSAR-COHN. Introduction to Modern Scientific 
Chemistry. In the form of popular lectures suited for 
University Extension students and general readers. 
Translated from the Second German Edition by M. M. 
Pattison Muir. Illus. i2mo. cloth. 356 pp. $2.00 

LETTS, E. A. Some Fundamental Problems in Chemis- 
try : Old and New. 44 illustrations. 8vo. cloth. 236 
pp. net, $2.00 

LUNGE, GEORGE. Technical Methods of Chemical 
Analysis. Translated from the Second German Edition 
by Charles A. Keane, with the collaboration of eminent 
experts. Complete in three volumes. Six parts. 448 
illustrations. 6>^x9>4. cloth. 3494 pp. net, $48.00 
Vol. I. (in two parts). 201 illustrations. 6^x91/2. 
cloth. 1024 pp. net, $15.00 

Vol. II. (in two parts). 149 illustrations. 6^^x9^. 
cloth. 1294 pp. net, $18.00 

Vol. III. (in two parts). 98 illustrations. 6j4 ^9^. 
cloth. 1 1 74 pp. net, $18.00 

Technical Chemists' Handbook. Tables and meth- 
ods of analysis for manufacturers of inorganic chemi- 
cal products. Illus. i2mo. leather. 276 pp. net, $3.50 

Coal, Tar and Ammonia. Fourth and Enlaro-ed Edi- 



tion. In two volumes, not sold separately. 305 illus- 
trations. 8vo. cloth. 1210 pp. net, $15.00 
— The Manufacture of Sulphuric Acid and Alkali. 



A theoretical and practical treatise. 



LIST OF CHEMICAL BOOKS 



Vol. I. Sulp'huric Acid. Fourth Edition, enlarged. 
In three parts, not sold separately. 543 illustrations. 
8vo. cloth. 1665 pp. net, $18.00 

Vol. II. Sulphate of- Soda, Hydrochloric Acid, Leblanc 
Soda. Third Edition, much enlarged.. In two parts, 
not sold separately. 335 illustrations. 8vo. cloth. 
1044 pp. net, $15.00 

Vol. III. Ammonia Soda. Various Processes of Al- 
kali-making, and the Chlorine Industry. 181 illus- 
trations. 8vo. cloth. 784 pp. net, $10.00 
Vol. IV. Electrolytical Methods. In Press. 

Technical Gas Analysis. 143 illustrations. 6x9. 



cloth. 4^2 pp. net, $4.00 

McINTOSH, JOHN G. The Manufacture of Varnish and 
Kindred Industries. lUus. 8vo. cloth. In 3 volumes. 
Vol. I. Oil Crushing, Refining and Boiling; Manu- 
facture of Linoleum ; Printing and Lithographic Inks ; 
India Rubber Substitutes. 29 illus. 160 pp. net, $3.50 
Vol. II. Varnish Materials and Oil Varnish Making. 
66 illus. 216 pp. net, $4.00 

Vol. III. Spirit Varnishes and Varnish Materials. 
64 illus. 492 pp. net, $4.50 

MARTIN, G. Triumphs and Wonders of Modern Chem- 
istry. A popular treatise on modern chemistry and 
its marvels written in non-technical language. y6 il- 
lustrations. i2mo. cloth. 358 pp. net, $2.00 

MELICK, CHARLES W. Dairy Laboratory Guide. 52 
illustrations. i2mG. cloth. 135 pp. net, $1.25 

MERCK, E. Chemical Reagents : Their Purity and Tests. 
Second Edition, revised. 6x9. cloth. 210 pp. $1.00 

MIERZINSKI, S. The Waterproofing of Fabrics. Trans- 
lated from the German by A. Morris and H. Robson. 
Second Edition, revised and enlarged. 29 illustrations. 
5x7^4. 140 pp. net, $2.50 



lo D. VAN NOSTRAND COMPANY'S 

MITCHELL, C. A. Mineral and Aerated Waters, iii 
illustrations. 8vo. cloth. 244 pp. net, $3.00 

MITCHELL, C. A., and PRIDEAUX, R. M. Fibres Used 
in Textile and Allied Industries. 66 illustrations. 
8vo. cloth. 208 pp. net, $3.00 

MTJNBY, A. E. Introduction to the Chemistry and 
Physics of Building Materials. lUus. 8vo. cloth. 365 
pp. (Van Nostrand's Westminster Series.) net, $2.00 

MURRAY, J. A. Soils and Manures. 33 illustrations. 
8vo. cloth. 367 pp. (Van Nostrand's Westminster 
Series.) net, $2.00 

NAQUET, A. Legal Chemistry. A guide to the detec- 
tion of poisons as applied to chemical jurisprudence. 
Translated, with additions, from the French, by J. P. 
Battershall. Second Edition, revised with additions. 
i2mo. cloth. 190 pp. $2.00 

NEAVE, G. B., and HEILBRON, I. M. The Identifica- 
tion of Organic Compounds. i2mo. cloth, iii pp. 

net, $1.25 

NORTH, H. B. Laboratory Experiments in General 
Chemistry. Second Edition, revised. 36 illustrations. 
5^x7^. cloth. 212 pp. net, $1.00 

OLSEN, J. C. A Textbook of duantitative Chemical 
Analysis by Gravimetric and Gasometric Methods. 
Including 74 laboratory exercises giving the analysis 
of pure salts, alloys, minerals and technical products. 
Fourth Edition, revised and enlarged. 74 illustrations; 
8vo. cloth., 576 pp. net, $4.00 

PAKES, W. C. G., and NANKIVELL, A. T. The Science 
of Hygiene. A text-book of laboratory practice. 80 
illustrations. i2mo. cloth. 175 pp. net, $1.75 

PARRY, ERNEST J. The Chemistry of Essential Oils 
and Artificial Perfumes. Second Edition, thoroughly 
revised and greatly enlarged. Illustrated. 8vo. cloth. 
554 pp. net, $5.00 



LIST OF CHEMICAL BOOKS ii 

Food and Drugs. In 2 volumes, Illus. 8vo. cloth. 

Vol. I. The Analysis of Food and Drugs (Chemical 
and Microscopical). 59 illus. 724 pp. net, $7.50 

Vol. II. The Sale of Food and Drugs Acts, 1873- 
1907. 184 pp. net, $3.00 

PARTINaTON, JAMES R. A Text-book of Thermo- 
dynamics (with special reference to Chemistry). 91 
diagrams. 8vo. cloth. 550 pp. .let, $4.00 

^ — Higher Mathematics for Chemical Students. 44 
diagrams. i2mo. cloth. 272 pp. net, $2.00 

PERKIN, F. M. Practical Methods of Inorganic Chem- 
istry. Illustrated. i2mo. cloth. 152 pp. net, $1.00 

PERKIN, F. M., and JAGGERS, E. M. Textbook of Ele- 
mentary Chemistry, yy illustrations. 4^ x 7. cloth. 
342 pp. net, $1.00 

PLATTNER'S Manual of Qualitative aiid auantitative 
Analysis with the Blowpipe. Eighth Edition, revised. 
Translated by Henry B. Cornwall, assisted by John 
H. Caswell, from the Sixth German Edition, by Fried- 
rich Kolbeck. 87 ill. 8vo. cloth. 463 pp. net, $4.00 

POLLEYN, F. Dressings and Finishings for Textile 
Fabrics and Their Application. Translated from the 
Third German Edition by Chas. Salter. 60 illustra- 
tions. 8vo. cloth. 279 pp. net, $3.00 

POPE, F. G. Modern Research in Organic Chemistry. 
261 diagrams. i2mo. cloth. 336 pp. net, $2.25 

PORRITT, B. D. The Chemistry of Rubber. 5x7><. 
cloth. 100 pp., (Van Nostrand's Chemical Mono- 
graphs.) net, $0.75 

POTTS, HAROLD E. Chemistry of the Rubber Industry. 
8vo. cloth. 163 pp. net, $2.00 

PRESCOTT, A. B. Organic Analysis. A manual of the 
descriptive and analytical chemistry of certain carbon 
compounds in common use. Sixth Edition. Illus- 
trated. 8vo. cloth. 533 pp. $5.00 



12 D. VAN NOSTRAND COMPANY'S 

PRESCOTT, A. B., and JOHNSON, 0. C. aualitative 

Chemical Analysis. Sixth Edition, revised and en- 
larged. 8vo. cloth. 439 pp. net, $3.50 

PRESCOTT, A. B., and SULLIVAN, E, C. First Book in 
Qualitative Chemistry. For studies of water solution 
and mass action. Eleventh Edition, entirely rewritten. 
i2mo. cloth. 150 pp. net, $1.50 

PRIDEAUX, E. B. R. Problems in Physical Chemistry 
with Practical Applications. 13 diagrams. 8vo. cloth. 
323 pp. net, $2.00 

PROST, E. Manual of Chemical Analysis. As applied 
to the assay of fuels, ores, metals, alloys, salts, and 
other mineral products. Translated from the original 
by J. C. Smith. Illus. 8vo. cloth. 300 pp. net, $4.50 

PYNCHON, T. R. Introduction to Chemical Physics. 
Third Edition, revised and enlarged. 269 illustrations. 
Svo. cloth. 575 pp. $3.00 

RICHARDS, W. A., and NORTH, H. B. A Manual of 
Cement Testing. For the use of engineers and chem- 
ists in colleges and in the field. 56 illustrations. 
i2mo. cloth. 147 pp. net, $1.50 

RIDEAL, S. Glue and G-lue Testing. Second Edition, 
revised and enlarged. 14 illustrations. 5^x8^. 
cloth. 194 pp. net, $4.00 

ROGtERS, ALLEN. A Laboratory Guide of Industrial 
Chemistry. Illustrated. 8vo. cloth. 170 pp. net, $1.50 

ROGERS. ALLEN (Editor). Industrial Chemistry. A 
manual for the student and manufacturer. Second 
Edition, thoroughly revised and enlarged. Written 
by a staff of eminent specialists. ^04 illustrations. 
65^x9^. cloth. 1026 pp. net, $5.00 

ROHLAND, PAUL. The Colloidal and Crystalloidal State 
of Matter. Translated by W. J. Britland and H. E. 
Potts. i2mo. cloth. 54 pp. net, $1.25 



LIST OF CHEMICAL BOOKS 13 

ROTH, W. A. Exercises in Physical Chemistry. Author- 
ized translation by A. T. Cameron. 49 illustrations. 
8vo. cloth. 208 pp, net, $2.00 

SCHERER, R. Casein: Its Preparation and Technical 
Utilization. Translated from the German by Charles 
Salter. Second Edition^ revised and enlarged. Il- 
lustrated. Svo. cloth. 196 pp. net, $3.00 

SCHIDROWITZ, P. Rubber. Its Production and Indus- 
trial Uses. Plates, 83 illus. 8vo. cloth. 320 pp. 

net, $5.00 

SCHWEIZER, V. Distillation of Resins, Resinate Lakes 
and Pigments. Illustrated. Svo. cloth. 183pp.net, $3.50 

SCOTT, W. W. Qualitative Chemical Analysis. A labo- 
ratory manual. Second Edition, thoroughly revised. 
Illus. Svo. cloth. iSo pp. net, $1.50 

SCUDDER, HEYWARD. Electrical Conductivity and 
Ionization Constants of Organic Compounds, 6x9. 
cloth. 575 pp. net, $3.00 

SEARLE, ALFRED B. Modern Brickmaking. 260 illus- 
trations. Svo. cloth. 449 pp. net, $5.00 

Cement, Concrete and Bricks. 113 illustrations. 

5>4 X814. cloth. 415 pp. net, $3.00 

SEIDELL, A. Solubilities of Inorganic and Organic Sub- 
stances. A handbook of the most reliable quantitative 
solubility determinations. Second Printing, corrected. 
Svo. cloth. 367 pp. net, $3.00 

SENTER, G. Outlines of Physical Chemistry. Second 
Edition, revised. Illus. i2mo. cloth. 401 pp. $1.75 

A Text-book of Inorganic Chemistry. 90 illustra- 
tions. i2mo. cloth. 595 pp. net, $1.75 

SEXTON, A. H. Fuel and Refractory Materials. Second 
Ed., revised. 104 illus. i2mo. cloth. 374 pp. net, $2.00 

■ Chemistry of the Materials of Engineering. Illus. 

i2mo. cloth. 344 pp. net, $2.50 



14 D. VAN NOSTRAND COMPANY'S 

SIMMONS, W. H., and MITCHELL, C. A. Edible Fats 
and Oils. Their composition, manufacture and analy- 
sis. Illustrated. 8vo. cloth. 164 pp. net, $3.00 

SINDALL, R. W. The Manufacture of Paper. 58 illus. 
8vo. cloth. 285 pp . (Van Nostrand's Westminster 
Series.) net, $2.00 

SINDALL, R. W., and BACON, W. N. The Testing of 
Wood Pulp. A practical handbook for the pulp and 
paper trades. Illus. 8vo. cloth. 150 pp. net, $2.50 

SMITH, J. C. The Manufacture of Paint. A manual for 
paint manufacturers, merchants and painters. Second 
Edition, revised and enlarged. 80 illustrations. SV^ ^ 
854. cloth. 286 pp. net, $3.50 

SMITH, W. The Chemistry of Hat Manufacturing. 
Revised and edited by Albert Shonk. Illustrated. 
i2mo. cloth. 132 pp. net, $3.00 

SOUTHCOMBE, J. E. Chemistry of the Oil Industries. 
Illus. 8vo. cloth. 209 pp. net, $3.00 

SPEYERS, C. L. Text-book of Physical Chemistry. 20 
illustrations. Svo. cloth. 230 pp. net, $2.25 

SPIEGEL, L. Chemical Constitution and Physiological 
Action. Translated by C. Luedeking and A. C. 
Boynton. In Press. 

STEVENS, H. P. Paper Mill Chemist. 67 illustrations. 
82 tables. i6mo. cloth. 280 pp. net, $2.50 

SUBBOBOUGH, J. J., and JAMES, J. C. Practical Or- 
ganic Chemistry. 92 illustrations. i2mo. cloth. 
394 pp. net, $2.00 

TERR¥, H. L. India Rubber and Its Manufacture. 
18 illustrations. 8vo. cloth. 303 pp. (Van Nos- 
trand's Westminster Series.) net, $2.00 

TITHERLEY, A. W. Laboratory Course of Organic 
Chemistry, Including Qualitative Organic Analysis. 
Illustrated. 8vo. cloth. 235 pp. net, $2.00 

TOCH, M. Chemistry and Technology of Mixed Paints. 
Nezu Edition. In Preparation. 



LIST OF CHEMICAL BOOKS 15 

TOCH, M, Materials for Permanent Painting. A manual 
for manufacturers, art dealers, artists, and collectors. 
With full-page plates. Illustrated. i2mo. cloth. 
208 pp. net, $2.00 

TUCKER, J. H. A Manual of Sugar Analysis. Sixth 
Edition. 43 illustrations. 8vo. cloth. 353 pp. $3.50 

UNDERWOOD, N., and SULLIVAN, T. V. Chemistry and 
Technology of Printing Inks. 9 illustrations. 6x9. 
cloth. 145 pp. net, $3.00 

VAN NOSTRAND'S Chemical Annual. Edited by John 
C. Olsen and Alfred Melhado. A handbook of useful 
data for analytical manufacturing and investigating 
chemists and chemical students. Third Issue, enlarged. 
5x7>^. leather. 683 pp. net, $2.50 

VINCENT, C. Ammonia and Its Compounds. Their 
manufacture and uses. Translated from the French 
by M. J. Salter. 32 ill. 8vo. cloth. 113 pp. net, $2.00 

VON GEORGIEVICS, G. Chemical Technology of Textile 
Fibres. Translated from the German by Charles 
Salter. 47 illustrations. 8vo. cloth. 320 pp. net, $4.50 

Chemistry of Dyestuffs. Translated from the Sec- 
ond German Edition by Charles Salter. 8vo. cloth. 
412 pp. net, $4.50 

VOSMAER, A. Ozone, Its Manufacture and Uses. 

In Press. 
WADMORE, J. M. Elementary Chemical Theory. Illus. 

T2mo. cloth. 286 pp. net, $1.50 

WALKER, JAMES. - Organic Chemistry for Students of 

Medicine. Illus. 6x9. cloth. 328 pp. net, $2.50 
WALSH, J. J. The Chemistry of Mining and Mine 

Ventilation. In Press. 

WARNES, A. R. Coal Tar Distillation and Working Up 

of Tar Products. 6y illustrations. 5^x8^. cloth. 

197 pp. net, $2.50 



i6 LIST OF CHEMICAL BOOKS 

WHITE, C. H. Methods in Metallurgical Analysis. io6 
illustrations. 5x71^. cloth. 365 pp. net, $2.50 

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2T4M— 4— 15 



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